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1
Chapter 1
1.1
ALGEBRAIC EXPRESSIONS
Introduction
An algebraic expression is a constant, a variable, or any combination of
constants and variables made by addition, subtraction, multiplication, division, or
extraction of roots.
Examples:
x2 - 2x + 5;
x2 2 x
xy y 2
;
3y
When an algebraic expression is written as the sum of other expressions, each
expression forming the sum is called a term. In the expression x/y + y2 - 5x + 4, the
terms are x/y, y2, - 5x, 4. When the variable factors of two terms are exactly alike,
the terms are said to be like terms or similar terms. Hence 2x2 and -3x2 are like
terms, while 5y and 4y2 are not. Algebraic expressions are added and subtracted
by combining like terms. Like terms are added by adding their coefficients.
A monomial is an algebraic expression which is composed of one term.
Example: 3x4; -5xy ;
3xy2 ; -y2zx
The constant of a monomial is called the numerical coefficient. Hence, 3 is the
numerical coefficient of 3x4, -5 is the numerical coefficient of -5xy, and -1 is the
numerical coefficient of -y2zx.
A binomial is an algebraic sum or difference of two terms.
A trinomial is an algebraic expression consisting of three terms.
A multinomial is an algebraic expression consisting of a sum of any number of
terms.
A polynomial is an algebraic expression, which may be written as a sum of terms
expressed as a product of a real number and non-negative integral powers of
variables.
Example: x2 + 4x - 3; 2x2 - 3xy;
The largest exponent for a given variable is called the degree of the polynomial in that
variable. Thus 4x3 - 3x2y + y2 is a polynomial of degree 3 in x and of degree 2 in y.
2
Other examples:
a ,
2b
and
2ab
------3bc
are monomials
3x2 - 2y
is a binomial
2a + 3b – 4
is a trinomial
3b2 4ab
a3
2a + ------ - ----- + ---2bc 2bc
2
is a multinomial
2x4y + 2x3y - 4x2y2 + 2xy3 - 4y2 is a polynomial
1.2
Grouping Algebraic Expressions
The following symbols are commonly used in mathematical statements to
indicate certain operations and to group parts of the expression together.
(
[
{
) ------- parenthesis
] ------- bracket
} ------- brace
The presence of certain symbols of grouping in algebraic expressions indicates that
certain operations are to be performed in a specified order. Frequently, it is desirable to
obtain an expression, which is equal to the given one that does not contain any of the
symbols anymore. The procedure to accomplish this is called removing the symbols of
grouping.
Steps in removing symbols of grouping:
1.
2.
3.
4.
If a compound symbol of grouping exists, start removing the innermost symbol.
If a term (numerical coefficient, literal coefficient or both) appears before a
symbol of grouping, multiply this term to every term inside the symbol.
a. If a positive (+) sign preceded a symbol, the symbol may be removed without
any changes done on the enclosed terms.
b. If a negative (-) sign proceeds a symbol, the symbols may be removed and
change the sign of every enclosed term.
Simplify the expression by combining like terms.
Example:
1.
a - b + 2(3a - 4b) - (-3a + 8b)
= a - b + 6a - 8b + 3a - 8b
= a + 6a + 3a - b - 8b - 8b
= 10a - 17b
3
2.
3x - {4x - 2y - 3[4x - 3y + z -(2x - 4y + 3z) - 2x] + 4z} - 3y
= 3x - {4x - 2y - 3[4x - 3y + z - 2x + 4y - 3z - 2x] + 4z} - 3y
= 3x - {4x - 2y - 12x + 9y - 3z + 6x - 12y + 9z + 6x + 4z} - 3y
= 3x - 4x + 2y + 12x - 9y + 3z - 6x + 12y - 9z - 6x - 4z - 3y
= 3x - 4x + 12x - 6x - 6x + 2y - 9y + 12y - 3y + 3z - 9z - 4z
= 15x - 16x + 14y - 12y + 3z - 13z
= -x + 2y - 10z
Exercises: Remove the symbols of grouping and simplify the following expressions.
1.
[5x + 2y - (3x + 5y) + (4x - 2y) + 4x - 3y]
2.
w - [2x - 3y - (2w + 4x - 3y) - 2w] - (x - y)
3.
[3x - 14 + {(x + 3y - 3y) - (4x - 6y) + 12x} + 3x + 5y]
4.
8x + 9y - 2{x + 5y - (4x - y) - (3x + 2y) - 7x} + 8x - 8y
5.
3a - (2b - 3c) - [a - 2b + (c + 3a - b) + c] - (a + 2b) + c
6.
3[2x - 4(x + y)] - 5{x2 - [3y - x(x - y)]}
7.
15x + 2y - 3{3x + 5y + (2x + 5) - (3x + 4y) + 17x} + 12y + 4
8.
2u - {3v - w + [4u - (2v - 3w)] - 3u} + v
9.
-[3x + 5y - (2x + 2y) + (5x - 2y) + 17x + 4y + 4]
10. 4x - 3{x + 2[4x - 5(x - 3)+ 5] - x}
Enclosing Algebraic Expressions with a symbol of grouping
To enclose some terms in an algebraic expression, change the signs of the terms
to be enclosed if to be preceded by a minus sign and retain the signs if to be preceded
by a plus sign.
Examples:
Enclose the last three terms in a symbol of grouping preceded by a) plus sign b)
minus sign.
1.
2a + 3b - 5a2 + 4b2
a.
2a + (3b - 5a2 + 4b)2
b.
2a - (-3b + 5a2 - 4b2)
4
2.
-5x2 + 7x3 - 2x + x4
a.
-5x2 + (7x3 - 2x + x4)
b.
-5x2 - (-7x3 + 2x - x4)
Exercise: Enclose the last three terms in a symbol of grouping preceded by a) plus
sign b) minus sign.
1.
a2 - b2 + 2bc - c2
4.
9x2 - 9y2 - 6xy - x2
2.
4x2 - 4y2 - 4y-1
5.
ax - by + bx - ay
3.
16 - x2 + 2xy - y2
1.3
Laws of Exponents
The following are the fundamental laws on exponents:
1. ao
= 1
2.
aman
=
a(m+n)
3.
(am)n
=
amn
4.
(ab)n
=
a n bn
5.
a
[----]n
b
=
an
---bn
6.
am
---an
=
a(m-n)
7. a-n
=
1
---an
8. a1/n
=
n%a
9. am/n
= (a1/n)m
=
am/n
= (am)1/n
1
= ----am/n
=
10. a-m/n
a
m
n
n
am
or
5
Examples:
1.
2.
(8a)o
a2a3
=
1
=
a(2+3)
= a5
= b12
3.
(b6)2
=
b6?2
4.
(x3y)2
=
x(3.2) y2 =
5.
x2
x 2 ( 3) x 6
3
y 3( 3) y 9
y
x6 y2
3
6.
m9
m ( 9 6 ) m3
m6
7. a 3
1
a3
8. x1/2
= %x
9. x2/3
x2/3
= (x1/3)2 = [3%x]2 or
= (x2)1/3 = 3%x2
10. p-2/3
1
= ---p2/3
In simplifying expressions with exponents, simply apply the laws on exponents and
then, if possible, reduce the resulting expression to its lowest terms. Below are more
examples combining the different laws or properties of exponents.
p3
1. ---p5
1
-----p(5-3)
=
1
= ---p2
2x 3 y 4 3y 2 z 3 2 4 x 12 y16 32 y 4 z 6
2. 3
2 4
2
12
z 2x z
2 x
=
32 24 x12 y 20z 6
22 x 4 z12
6
4
3.
=
3 2 2 2 x 8 y 20
z6
=
36x 9 y 20
z6
12x ( 2 a 2 ) 1
4 1
2x ( 2a 2) (a 2) 2a
(a 2)
2a
2x
6x
2x
2
4 1
= 2x 2 a 2a 2 2a
2x
4
= 2x
a
1
2x 2a
2
1
= 2 4 x 4a 2 4a
2 x
=
2 4 x 4a
2 2 x 4a
= 4
3
4
3
4 r 2s 3
9s 2 t 4
4r 2s 3 8r 3
4. 4 3 4 2 4
3t
8r
3t 9s t
43 r 6s9 84 r12
= 3 12 4 8 16
3 t 9 s t
=
84 43 r18s9
9 433 s8 t 28
84 43 r18s
= 4 3 28
93t
4
2
2
7
c
5.
( a 3)
m( a 2 )
c3a m 6
3
=
=
c
3( a 3)
m 3( a 2 )
c 3a m 6
c 3a 9 m 3a 6
c 3a m 6
= c(3a+9)-(3a)m(3a+6)-(6)
= c9 m3a
6.
2 3 a 2 bc 1
4 2 xy 3 z 4
a 2 b 3
7. 2 1 2
2 a b
=
4 2 by 3
(2 2) 2 by 3
2 3 a 2 cxz 4
2 3 a 2 cxz 4
=
2 2 2 2 by 3
2 3 a 2 cxz 4
=
2by 3
a 2 cxz 4
2 2 ab 3 b 2
=
2
a
3
22 b5
a
2 6 b 15
a3
=
6 15
a 3
2 b
=
a3
64b15
3
8
Exercise: Simplify and reduce the following exponential expressions:
1.
64r 3s 6 p 5
56r 5 s 4 p 3
2.
7b 4 c 3 16b 3 d 3
8d 2 21c 4 d 5
3.
7 p 2 q 4 12p 3s 7
2 3
5 3 4
8r s 21p q s
9.
3
4.
v
w ( 2 z 1)
v ( 6 z 4) w ( 2 z 4)
3
1.4
11.
4r 2 s 3 9s 2 t 4
4 3
3t 8r
3
4
5.
24b 2 x 5
12b 2 x 3
3 2
5 4
a y
a y
6.
3 2 u 3 v 2 w 2
6 1 u 4 v 1 w 3
7.
9 1 m 2 g 2
3 3 3
6 m g
8.
10.
3x 2 y 3 28y 3
7 y 4 z 2 33x 4 z 3
3
( 3 z 1)
72m 8 e 7 c 4
54m 7 e 6 c 5
12.
2
( 2 n 3)
z ( n 2)
y ( n 9) z (3n 1)
4
3
13.
3d 3 a 2 6d 2 a 4
4 3 3 2
2b c 4b c
14.
5 3 x 2 y 4 z 2
10 2 x 3 y 3 z 4
15.
8 2 x 3 y 2
3 4 4
4 x t
16.
a 2 b 2
a 2 b 2
2
b 4 c 3
b 5
y
4
2
Operations on Algebraic Expressions
1.4.1 Addition/Subtraction of Algebraic Expressions
In a particular term consisting of two or more factors, any one of the factors or the
product of any set of the factors may be called the coefficient of the product of the other
factors. For example, in the term 2x2y, 2 is the coefficient of x2y, 2x2 is the coefficient of
y and 2y is the coefficient of x2. The coefficients are used in determining the sum or
difference of two algebraic expressions by combining similar terms.
9
Examples:
1.
-5b + b + 6b = -5b + 7b
= 2b
2.
2a - 3a + 6b + 4b - 7c + 9c = -a + 10b + 2c
3.
(-4a + 2b - 5c) + (2a - 3c + 4b) + (6c + 7a - 8b) =
= -4a + 2b - 5c + 2a - 3c + 4b + 6c + 7a - 8b
= -4a + 2a + 7a + 2b + 4b - 8b - 5c - 3c + 6c
= -4a + 9a + 6b - 8b - 8c + 6c
= 5a - 2b - 2c
4.
(4a - 3b + 3c) - (-3a + 2b - 5c) = 4a - 3b + 3c + 3a - 2b + 5c
= 4a + 3a - 3b - 2b + 3c + 5c
= 7a - 5b + 8c
5.
(z2 - 4z - 3) + (2z2 + 3z + 2) - (z2 + 3z + 5)
= z2 - 4z - 3 + 2z2 + 3z + 2 - z2 - 3z - 5
= z2 + 2z2 - z2 - 4z + 3z - 3z - 3 + 2 - 5
= 3z2 - z2 - 4z - 8 + 2
= 2z2 - 4z - 6
Exercise: Determine the sum of the following algebraic expressions.
1.
(6p2 - 3q2) + 4q2 - 7r2 + (-2p2 + 5r2)
2.
7r2 + 5rs + 2s2 - (4r2 - 2rs + 6s2)
3.
2r - 3s + 5t + (3r - 5s + 7t) + 5r + 7s - 4t
4.
Subtract 5x2 - x + 7 from the sum of x3 + x2 + x - 2 and 2x3 - 3x2 + 2x - 5
5.
(4xy - 5yz + 6xz) + (5xy - 7yz - 2xz) + [-2xy + 3yz + 5xz]
6.
3s2 - 2s + 5 - [(2s2 - 5s + 3) + (4s2 + 3s - 4)]
7.
4x5 - 2x3 + 5 + (3x4 - 2x2 - 2) + (-2x5 + x2 + 3)
8.
Subtract 6x2 + 3y2 -3z2 from the sum of 3x2 - 7y2 + 2z2 and 4x2 + 5y2 - 4z2
9.
(8a3 - 5a2 + 3a) + (7a2 - 4a3 - 2a) + 6a + 3a2 - 5a3
10. 4x3 + 2xy2 - y3 - (2x2y - 3xy2 - 3x3)
10
1.4.2 Multiplication of Algebraic Expressions
The product of polynomials is obtained by the use of the commutative, associative
and distributive axioms, law of signs and the law of exponents in multiplication.
Examples:
1. 3a4 . 8a3
= 24a7
2. 2xy3 (-3x2y)
= -6x3y4
3. (4a2b)3 (5a4b2) = (43a6b3) (5a4b2) = 43 . 5 a10b5
= 320a10b5
4. 3x2yz3 (-2x3y2z) (6xy4z5) = -36x6y7z9
5. 3ab (2a - 4b + 7a2b)
= 6a2b - 12ab2 + 21a3b2
6. (3x2y - 6xy2 - 8y3) (-5x3y2) = -15x5y3 + 30x4y4 + 40x3y5
7. 4ab2 (3a2b - 2a3 + 5ab3) - (3ab - 4a2 + 15b3) 2a2b2
= (12a3b3 - 8a4b2 + 20a2b5) - (6a3b3 - 8a4b2 +30a2b5)
= 12a3b3 - 8a4b2 + 20a2b5 - 6a3b3 + 8a4b2 - 30a2b5
= 6a3b3 - 10a2b5
8. (3a + 2b) (a - 3b) = 3a2 - 9ab + 2ab - 6b2
= 3a2 - 7ab - 6b2
or:
3a + 2b
a - 3b
--------------3a2 + 2ab
+
- 9ab - 6b2
----------------------3a2 - 7ab - 6b2
9. (p -4q + r) (3p + q - 2r) = 3p2 + pq - 2pr - 12pq - 4q2 + 8qr + 3pr +
11
or:
p - 4q + r
3p + q - 2r
--------------------3p2 - 12pq + 3pr
+ pq
- 4q2 + qr
- 2pr + 8qr - 2r2
-----------------------------------------3p2 - 11pq + pr -4q2 + 9qr - 2r2
Exercise: Determine the product of the following algebraic expressions:
1. 2r3s2 (-4rt3) (-s3t3)
2. -4bc3 (3bc2 - 6b4c)
3. (-4c4d2 - 3c2d4) (-2c3d)
4. 3r2s3t (-3rst2 + 2r3t - r2s3)
5. 3x3y4z2 (2x4y4z4 - 4x2y3z6) - 5x4y5z3 (3x3y3z3 - 2xy2z5)
6. (3b + 4c) (2b - 3c)
7. (x + 3y) (3x - y)
8. (m + 2n) (m2 + 2mn - n2)
9. (x + 2y - z) (2x + y - 2z)
10. (7x3 - 2x2 + 3x - 5) (2x2 +3x - 4)
1.4.3 Division of Algebraic Expressions
Obtaining the quotients of algebraic expressions requires knowledge of the laws on
signs and exponents. Also, the following theorems should be recalled and familiarized:
12
1 1
1
-- . -- = ---a b
ab
(a and b ≠ 0)
a
1
1
-- = a . -- = -- . a
b
b
b
(b ≠ 0)
cd
c d
c d
-- = -- . -- = - . ab
a b
b a
(a and b ≠ 0)
a+b+c a b c
----------- = - + - + d
d d d
(d ≠ 0)
1.4.3.1 Division by Monomials
Examples:
1.
a 3b5
ab 2
2 3
a b
2.
24a 4 b10 c 2
4.
8b 3 c 2
4 7
3a b
12p 5 q 9 r 7
3.
3p 2 q 3 r 5
3 6 2
4p q r
5.
6x 8 y 6 z 3
2x 4 y 3 z
4 3 2
3x y z
6x 4 4x 3 y 3 3x 2 y 2 2x 2 6x 4 4x 3 y 3 3x 2 y 2 2x 2
2
2
3x 2
3x
3x 2
3x 2
3x
= 2x 2
4
2
xy 2 y 2
3
3
1.4.3.2 Division by Polynomials
The degree of a polynomial in any variable is the greatest exponent of that variable
in the polynomial. For example, the polynomial 3x4 + 2x3y + 5x2y2 + xy3 is of degree 4
in x and 3 in y, and degree 4 in x and y.
In dividing a polynomial by another polynomial, the following steps are
recommended:
13
1. Arrange the terms in the dividend and divisor in the order of descending
powers of a letter that appear in each polynomial.
2. Divide the first term in the dividend by the first term in the divisor to get the
first term in the quotient.
3. Multiply the first term in the quotient by the divisor and subtract the product
from the dividend.
4. Treat the remainder obtained in step 3 as a new dividend and repeat steps 2
and 3.
5. Continue this process until a remainder is obtained that is of lower degree
than the divisor in the letter chosen in step 1 as the basis for the
arrangement.
Examples:
1.
Divide 3x3 - 4x2y + 5xy2 + 6y3 by x2 - 2xy + 3y2
(divisor)
2.
3x + 2y (quotient)
___________________________
x2 - 2xy + 3y2 ) 3x3 - 4x2y + 5xy2 + 6y3
(dividend)
3x3 - 6x2y + 9xy2
------------------------2x2y - 4xy2 + 6y3
2x2y - 4xy2 + 6y3
------------------0
0
0
Divide 5x3 - 14x + 3 by x - 2
5x3 + 0x2 - 14x + 3
[ x-2
3
2
5x - 10x
----------------------------------------------------[ 5x2 + 10x + 6
10x2 - 14x + 3
----------------------10x2 - 20x
-------------------------6x + 3
6x - 12
----------15 (remainder)
The quotient is:
5x2
15
+ 10x + 6 + ---x-2
14
1.4.3.3 Synthetic Division
The process of division for polynomials in x (or any other single letter) can be
greatly simplified if the divisor is in the form of "x - a", x being any letter and a being any
integer. The following steps summarize the process of synthetic division.
1. Determine if both the dividend and divisor contain the same singular literal
coefficient.
2. Determine if the divisor is in the form of "x - a".
3. Find the value of "a" by equating the divisor to x - a.
4. Arrange on a line (in order of descending power) the numerical coefficients,
inserting zero for the coefficient of any missing power of x(or any other
letter). Write the value of "a" (the new divisor) to the right of the numerical
coefficients.
5. Bring down the first numerical coefficient to the first position on the third line.
Multiply this first numerical coefficient by "a" then write the product in the
second line under the second numerical coefficient of the dividend. The
sum of this product and the second numerical coefficient is placed
underneath in the third line. Multiply this sum by "a" and add the product
to the next numerical coefficient of the dividend, again writing the new sum
of the third line, and so on, until a product has been added to the last
numerical coefficient of the dividend. The last sum in the third line
represents the remainder.
6. The numbers in the third line are the numerical coefficients of x(or any other
letter) in the quotient, arranged in descending order. The quotient is a
polynomial one degree less than the dividend.
Example:
Divide 5x3 - 14x + 3 by x - 2 using synthetic division
The divisor is of the form x - a where a = 2
Thus, using the numerical coefficients of the given dividend and the value of "a" as
the divisor:
15
5
0
-14 + 3
[ 2___
+
10 +20 + 12
----------------------------5 + 10 + 6 + 15
Hence, the quotient will be 5x2 + 10x + 6 remainder 15 or
15
5x2 + 10x + 6 + ---x-2
Exercises:
1.
15a 12 b 9 c 7
5a 8 b 7 c 5
2.
28r 8 s11t 6
7r 6 s 7 t 2
3.
48b13c 7 d 9
4b 8 c 5 d 6
4.
9x 7 y 7 z 8 15x 11 y 6 z 7 21x 2 y 9 z 3
3x 7 y 6 z 3
5.
25r 9 s 9 t 7 35r 2 s10 t 8 35a 11b 8 c 6
5r 2 s 3 t 5
6.
21a 10 b 9 c 7 14a 9 b 7 c 8 35a 11b 8 c 6
7a 8 b 6 c 5
7.
2a2 + 5a – 3 a + 3
8.
6x2 – 5xy – 6y2 3x + 2y
9.
5c4 – 13c3d + 4c2d2 – 21cd3 – 10d4 5c + 2d
10. x4 + 3x3 – 3x – 2 x2 – 5
11. 3x4 – x3y + 4xy3 – 2y4 x2 – xy + y2
12. 3x4 – 7x – 20 x + 2
13. x4 – 5x3 + x2 - 6 x – 1
16
1.5
Special Products and Factoring
You will recall that, when two or more numbers or algebraic expressions are
multiplied to form a product, they are called the factors of the product. When we start
with a product and break it up into its factors, the process is called factoring.
Since multiplication and factoring are so closely related, we shall include in this
section some exercises in certain types of multiplication and emphasize certain short
cut methods by which some types of products may be found.
1.5.1 The Product of a Monomial and a Polynomial
For the multiplication of a polynomial by a monomial we have the
Rule: multiply each term of the polynomial by the monomial and write the resulting
terms in succession with their proper signs.
Examples:
1.
a(b + x) = ab + ax
2.
3x(x2 - 2) = 3x3 - 6x
3.
-2xy(3x3 - x2 + 7x - 5) = -6x4y + 2x3y - 14x2y + 10xy
4.
am(3a3m - 5am + a - 11) = 3a4m - 5a2m + am+1 - 11am
1.5.1.1 Finding a Common Factor
When a factor is contained in every term of an algebraic expression, it is called
a common factor. A common factor may be either a monomial or a polynomial. A
polynomial can be factored by expressing it as the product of the common factor
and the sum of the quotients obtained by dividing each term of the polynomial by
that common factor.
Examples:
1. ab + ac - ad = a(b + c - d)
2. a(x2 + y2) - a(x2 - xy - y2)
=
=
=
=
a[(x2 + y2) - (x2 - xy - y2)]
a(x2 + y2 - x2 + xy + y2)
a(2y2 + xy)
ay(2y + x)
17
3. (a + b)(a - b) + 2(a + b)
= (a + b)[(a - b) + 2]
= (a + b)(a - b + 2)
4. (x - 1)(x + 2) - (x - 1)(2x - 3) = (x - 1)[(x + 2) - (2x - 3)]
= (x - 1)[x + 2 - 2x + 3]
= (x - 1)(5 - x)
Exercises:
For numbers 1-5, multiply the given expressions:
1.
2.
3.
4.
5.
-4a(1 - 2ab)
3x(5x3 - 2x2 - 5)
3/4(-16 + 8x - 32x2)
-2x2(6 - 4x - 7x3)
ah2(a - h3)
For numbers 6-10,factor the given expressions:
6.
7.
8.
9.
10.
6a3b + 3a2b2 - 18ab3
7x - 21y
3x3 - 6x2 + 9x
4x2 - 10x + 2
6x4 - 15x3 + 21x2
1.5.2 The Product of Binomials Having Dissimilar Terms
Multiply the binomials (2a - b)(x + 3y) gives us the product of 20x - bx + 6ay - 3by.
Since there are no similar terms in the product, we leave the answer in this form.
Examples:
1.
(m + 5)(m + 3n) = m2 + 3mn + 5m + 15n
2.
(12a + 10c)(x - 2y) = 12ax - 24ay + 10cx - 20cy
1.5.2.1 Factoring by Grouping
Expressions, which can be factored by this method, are easily recognizable by the
fact that there is always an even number of terms and the terms can be so arranged
that the numerical coefficients, disregarding signs, form a proportion. The terms in the
polynomials can be grouped in such a way that each group has a common factor.
18
Examples:
1. ax + bx - ay - by = (ax + bx) - (ay + by)
= x(a + b) - y(a + b)
= (a + b)[x - y]
2. 2ax + 4ay + 3x - 3z + 6y - 2az
= (2ax + 3x) + (4ay + 6y) - (2az + 3z)
= x(2a + 3) + 2y(2a + 3) - z(2a + 3)
= (x + 2y - z)(2a + 3)
3. 4x2 - 12xy + 9y2 + 4x - 6y - 3 = (4x2 - 12xy + 9y2) + (4x - 6y) - 3
= (2x - 3y)2 + 2(2x - 3y) - 3
= [(2x - 3y) + 3][(2x - 3y) - 1]
= [2x - 3y + 3][2x - 3y - 1]
Exercises:
For numbers 1-5, multiply the given expressions:
1. (x + 4)(x + 3)
2. (2x + 3)(4x + 5)
3. (5x + 7y)(6x - 5y)
4. (3w - 4r)(2w + 3r)
5. (3y - 2)(2 - 5y)
For numbers 6-10,factor the given expressions:
6. xy + x + 3y + 3
7. 2x2 + 4xy - x - 2y
8. 2x3 - 5x2 + 6x - 15
9. 3x2 - 12xy + 2x - 8y
10. xy - 2x + 4y - 8
1.5.3 The Product of Two Binomials Having Similar Corresponding Terms
The product of two binomials with similar corresponding terms can be obtained
by performing the following steps:
1.
2.
3.
Multiply the first terms of the two binomials to obtain the first term of the
product.
Add the products obtained by multiplying the first term in each binomial by the
second term in the other. This yields the second term of the product.
Multiply the second terms in the two binomials to get the third term of the
product.
19
Hence,
(ax + b)(cx + d) = acx2 + (bcx + adx) + bd
Examples:
1. (x + 2)(x + 5)
= (x)(x) + (2x + 5x) + (2)(5)
= x2 + 7x + 10
2. (3x + 4y)(2x - 3y) = (3x)(2x) + (8xy - 9xy) + (4y)(-3y)
= 6x2 - xy - 12y2
3. (2a - 5b)(4a + 3b) = (2a)(4a) + (-20ab + 6ab) + (-5b)(3b)
= 8a2 - 14ab - 15b2
4. (5m + 7n)(6m - 7n) = (5m)(6m) + (42mn - 35mn) + (7n)(-7n)
= 30m2 + 7mn - 49n2
1.5.3.1 Factoring a General Quadratic Trinomial
A trinomial of the type ax2 + bxy + cy2 (a, b, c are integers) is a quadratic trinomial
with integral coefficients. It is a product of two binomials. Factoring this type of
trinomial would require our knowledge in special products in obtaining the product of
two binomials. To recall:
acx2 + (ad + bc) xy + bdy2 = (ax + by)(cx + dy)
Examples:
1. 3x2 - 10xy - 8y2 = (3x + 2y)(x - 4y)
2. 12x2 - xy - 20y2 = (4x + 5y)(3x - 4y)
3. 6x4 + 7x2y2 - 3y4 = (3x2 - y2)(2x2 + 3y2)
4. 20m2 + 3mn - 35n2 = (4m - 5n)(5m + 7n)
5. (x + y)2 - (x + y) - 12 = [(x + y) - 4][(x + y) + 3]
20
You will find that you often will have to try several combinations of terms before
you find the right one. Just be sure you multiply the binomials to see if the product
agrees with the original trinomial.
Exercises:
For numbers 1-5, multiply the given expressions:
1. (x - 8)(x + 3)
2. (3r - 4s)(2r + 3s)
3. (6a - 2b)(3a + b)
4. (w + z)(3w - 2z)
5. (2c + 3d)(3c + 2d)
For numbers 6-10,factor the given expressions:
6. x2 - 2x - 8
7. 15x2 + 2x - 8
8. 12 - 7y + y2
9. 45x2 - 8xy + 4y2
10. 3y3 + 2y2 - 5y
1.5.4 The Square of a Binomial
The square of a binomial containing the sum (or the difference) of two terms is
equal to the square of the first term, plus (or minus) twice the product of the two terms,
plus the square of the second term.
Hence,
(x + y)2 = x2 + 2xy + y2
Examples:
1. (2a + 5b)2 = (2a)2 + 2(2a)(5b) + (5b)2
= 4a2 + 20ab + 25b2
2. (3x - 4y)2 = (3x)2 + 2(3x)(-4y) + (-4y)2
= 9x2 - 24xy + 16y2
3. (2x + 7y)2 = (2x)2 + 2(2x)(7y) + (7y)2
= 4x2 + 28xy + 49y2
4. (a - 3)2
= (a)2 + 2(a)(-3) + (-3)2
= a2 - 6a + 9
21
1.5.4.1 Factoring Perfect Square Trinomials
Some quadratic trinomials are said to be perfect square trinomials. A perfect
square trinomial is a polynomial wherein the first and the third terms are perfect squares
(making them both positive) while the middle term is twice the product of the positive
square roots of the first and third terms. Such trinomial is the square of a binomial
whose terms are the positive square roots of the first and third terms in the said perfect
square trinomial connected by the sign similar to that of the middle term in the trinomial.
Examples:
1. 4x2 - 12xy + 9y2 = (2x - 3y)2
2. 9a2 + 24ab + 16b2 = (3a + 4b)2
3. 9m2 - 30mn + 25n2 = (3m - 5n)2
4. 25 + 30(2x - 3y) + 9(2x - 3y)2 = [5 + 3(2x - 3y)]2
5. (2a - 3b)2 - 8(2a - 3b) + 16 = [(2a - 3b) - 4]2
Exercise:
For numbers 1-5, perform the indicated operation:
1.
2.
3.
4.
5.
(x2 + 2)2
(4ax2 - y)2
[2(x - y)]2
(x + 1/3)2
(2xy - 3y2)2
For numbers 6-10,factor the given expressions:
6.
7.
8.
9.
10.
x2 + 6x + 9
9x2 + 24x + 16
25x2 + 10x + 1
25x2 + 60x + 36
4a2 - 4a + 1
22
1.5.5 The Product of the Sum and Difference of the Same Binomials
The product of the sum and the difference of the same binomials is equal to the
difference of the square of each term.
Hence,
(a + b)(a – b) = a2 – b2
Examples:
1. (2x + 3y)(2x - 3y) = (2x)2 - (3y)2
= 4x2 - 9y2
2. (7x + 5y2)(7x - 5y2) = (7x)2 - (5y2)2
= 49x2 - 25y4
3. (3x + 4y)(3x - 4y) = (3x)2 - (4y)2
= 9x2 - 16y2
4. (8r2 + 3s5)(8r2 - 3s5) = (8r2)2 - (3s5)2
= 64r4 - 9s10
1.5.5.1 Factoring the Difference of Two Squares
The difference of two squares can be factored into the product of the sum and
difference of the square roots of the squares.
Hence:
a2 - b2 = (a + b)(a - b)
a2 – b2 = (a + b)(a – b)
Examples:
1. 4x4 - 9y2 = (2x2 + 3y)(2x2 - 3y)
2. (a + 3b)2 - 4 = [(a + 3b) + 2][(a + 3b) - 2]
3. x6 - (y + z)2 = [x3 + (y + z)[x3 - (y + z)]
23
4.
5.
2p
2p
4p 2
(9r t ) 2 (9r t ) (9r t )
2
q
q
q
a8 - b8 =
=
=
=
(a4 + b4)(a4 - b4)
(a4 + b4)[(a2 + b2)(a2 - b2)]
(a4 + b4)(a2 + b2)[(a + b)(a - b)]
(a4 + b4)(a2 + b2)(a + b)(a - b)
Exercises:
For numbers 1-5, multiply the given expressions:
1.
2.
3.
4.
5.
(cd - x)(cd + x)
(3r - 5s)(3r + 5s)
(5 - 2y)(5 + 2y)
(3x - 4z)(3x + 4z)
(ab - 2)(ab + 2)
For numbers 6-10,factor the given expressions:
6.
7.
8.
9.
10.
a2 - x2
16x2 - 81y2
16x4 - y4
49y2 - 3x2
p2q2 - 4r2
1.5.6 The Cube of a Binomial
The cube of a binomial containing the sum (or the difference) of two terms is equal
to the cube of the first term plus thrice the product of the square of the first term and the
second term plus thrice the product of the first term and the square of the second term
plus the cube of the second term.
Hence,
(x + y)3 = x3 + 3x2y + 3xy2 + y3
Examples:
1. (2x + 3y)3 = (2x)3 + 3(2x)2(3y) + 3(2x)(3y)2 + (3y)3
= 8x3 + (4x2)(9y) + (6x)(9y2) + 27y3
= 8x3 + 36x2y + 54xy2 + 27y3
24
2. (a - 2b)3 = (a)3 + 3(a)2(-2b) + 3(a)(-2b)2 + (-2b)3
= a3 - 6a2b + 3a(4b2) - 8b3
= a3 - 6a2b + 12ab2 - 8b3
Actually, the square or cube of a binomial is a part of the binomial theorem in which
the Pascal's Triangle is based. The Pascal's Triangle contains the numerical
coefficients of the terms of the product after a binomial expression is raised to a certain
exponent.
Exercises: Perform the indicated operation:
1.
2.
3.
4.
5.
(y - 3)3
(2x + 3y)3
(3x - 2)3
(2x2 - 5y2)3
(x - 1/3)3
1.5.7 The Square of a Polynomial
The square of a polynomial is equal to the sum of the square of each of its terms
plus twice the algebraic product of each term by every term that follows it in the
polynomial. This rule applies even if the polynomial has more than three terms.
1. (x + y + z)2 = [(x + y) + z]2
= (x + y)2 + 2(x + y)(z) + z2
= (x2 + 2xy + y2) + 2(xz + yz) + z2
= x2 + 2xy + y2 + 2xz + 2yz + z2
= x2 + y2 + z2 + 2xy + 2xz + 2yz
2. (2a- 3b - 5c)2 = [(2a - 3b) - 5c]2
= (2a - 3b)2 + 2(2a - 3b)(-5c) + (-5c)2
= (2a)2 + 2(2a)(-3b) + (-3b)2 + (4a - 6b)(-5c) + (-5c)2
= 4a2 - 12ab + 9b2 - 20ac + 30bc + 25c2
3. (a + b + c + d)2 = [(a + b + c) + d]2
= (a + b + c)2 + 2(a + b + c)(d) + d2
= [(a + b) + c]2 + 2(ad + bd + cd) + d2
= (a + b)2 + 2(a + b)(c) + c2 + 2(ad + bd + cd) + d2
= a2 + 2ab + b2 + 2(ac + bc) + c2 + 2(ad + bd + cd) + d2
= a2 + 2ab + b2 + 2ac + 2bc + c2 + 2ad + 2bd + 2cd + d2
= a2 + b2 + c2 + d2 + 2ab + 2ac + 2ad + 2bc + 2bd + 2cd
4. (2x + 3y - 2w - 4z)2 = (2x)2 + (3y)2 + (-2w)2 + (-4z)2 + 2(2x)(3y)
+ 2(2x)(-2w) + 2(2x)(-4z) + 2(3y)(-2w) + 2(3y)(-4z)+ 2(-2w)(-4z)
= 4x2 + 9y2 + 4w2 + 16z2 + 12xy - 8xw - 16xz - 12yw - 24yz + 16 wz
25
Exercises: Perform the indicated operation:
1.
2.
3.
(c + 2d - a)2
(2y + w - 3z)2
(2a + 2b + 2)2
1.5.8 Others Types of Factoring
1.5.8.1 Factoring the Sum or Difference of Two Cubes
A binomial expressed as the sum (or the difference) of the cubes of two numbers
has two factors. The first factor is a binomial containing the sum (or the difference) of
the cube roots of the two numbers. The second factor is a trinomial which contains the
square of the first cube root, minus (or plus) the product of the two cube roots plus the
square of the second cube root. Thus:
x3 + y3 = (x + y)(x2 – xy + y2
x3 – y3 = (x – y)(x2 + xy + y2)
Take note that the sign of the middle term in the second factor is opposite to the sign in
between the two terms in the first factor.
Examples:
1. a3 + 8
= (a + 2)[(a)2 - (a)(2) + (2)2]
= (a + 2)(a2 - 2a + 4)
2. 27a3 - 64b6 = (3a - 4b2)[(3a)2 + (3a)(4b2) + (-4b2)2
= (3a - 4b2)(9a2 + 12ab2 + 16b4)
3. 8x3 + 27y3
= (2x + 3y)[(2x)2 - (2x)(3y) + (3y)2]
= (2x + 3y)(4x2 - 6xy + 9y2)
4. a12 - 8b6
= (a4 - 2b2)[(a4)2 + (a4)(2b2) + (-2b2)2]
= (a4 - 2b2)(a8 + 2a4b2 + 4b4)
5. x6 - y6
= (x2 - y2)[(x2)2 + (x2)(y2) + (-y2)2]
= (x2 - y2)(x4 + x2y2 + y4)
= (x + y)(x - y)(x4 + x2y2 + y4)
26
Exercises: Factor the following expressions:
1.
2.
3.
4.
5.
a3 + b 3
x3 - 8y3
27x3 - 512y3
x6 - y3
x12 - y9
1.5.8.2 Factoring Binomials of the Form
an " bn where n is odd.
When n is an odd integer, one factor of an + bn (or an - bn) is a + b (or a - b). The
polynomial factor is in descending powers of a and ascending powers of b. When the
binomial factor is a + b, the signs in the polynomial factor alternate. When the binomial
factor is a - b, all the signs in the polynomial factor are positive.
Examples:
1.
a5 + b5 = (a + b)(a4 - a3b + a2b2 - ab3 + b4)
2.
a7 - 128 = (a - 2)(a6 + 2a5 + 4a4 + 8a3 + 16a2 + 32a + 64)
1.5.8.3 General Suggestions for Factoring
Usually a student finds it comparatively easy to factor a list of exercises, which are
all of one type form. A list of exercises involving all the types studied presents more
difficulty. The following suggestions for factoring a varied list of exercises will prove
helpful:
1.
First look for a common monomial factor (other than 1) and separate the
expression into its monomial factor and the corresponding polynomial factor.
2.
Study the polynomial factor after the monomial factors are removed. Determine to
which type of factoring it should be classed and then use the methods of factoring
applicable to that type.
3.
Continue with step 2 until all the factors obtained are prime factors.
27
1.5.8.4 Factoring Completely
An algebraic expression may have more than two factors. When two factors have
been found, each should be inspected to see whether further factoring is possible. With
proper groupings, an expression can also be factored applying the different types of
factoring we have already studied.
Examples:
1.
6a3b + 3a2b2 - 18ab3
= 3ab (2a2 + ab - 6b2) Common monomial factor
= 3ab (2a - 3b)(a + 2b) Quadratic trinomial
2.
4x3 - 12x2 - x + 3
= (4x3 - 12x2) - (x - 3) Grouping
= 4x2 (x - 3) - 1(x - 3)
= (4x2 - a)(x - 3)
= (2x + 1)(2x - 1)(x - 3) Difference of Squares
3.
a2 + ab - 2b2 + 2a - 2b
= (a2 + ab - 2b2) + (2a - 2b) Grouping
= (a + 2b)(a - b) + 2(a - b) Quadratic trinomial
= (a + 2b + 2)(a - b)
4.
4c2 - a2 + 2ab - b2
= 4c2 - (a2 - 2ab + b2) Perfect square trinomial
= 4c2 - (a - b)2
= [2c + (a - b)][2c - (a - b)] Difference of squares
= (2c + a - b)(2c - a + b)
Exercises 1. Determine the following special products.
1. (3x + 2)(c - 3)
2. (2s - 9)(3x + 7)
3. (6t + 5)(5t - 4)
4. (5a - 2b)2
5. (5r + 7s)2
6. (8p - 3q)2
7. (2a + b)3
8. (2x - 3y)3
9. (z + 3)3
10. (2m + 5)(2m - 5)
11. (2x + 9y2)(2x - 9y2)
x
12.
3
y x y
5 3 5
13. (y2 - 2y + 7)2
14. [(a2 - 5) + a][(a2 - 5) - a]
28
15. [2(x + y) - 3][3(x + y) + 5]
16. (x3 + x2 - 2x + 1)2
17. (a - 2b - c + 3d)2
18. (4w - 2x + 3y - z)2
Exercise 2. Factor the following algebraic expressions.
1.6
1.
b2 - 8bc + 12c2
2.
30u2 - 13up - 56p2
3.
10x2 + 7x(2y - 3z) - 12(2y - 3z)2
4.
9a2 + 6ab + b2
5.
25r2 - 30rs + 9s2
6.
16p2 + 56pq + 49q2
7.
36c8 - 49d6
8.
100u10 - 49v12
9.
144a16 - 25b14
10. k9p3 + 216
11.
125m9t6 - 8
12. 27x15y12 - 64
13.
3x2 - 9x - 30
14. 2a2 + 10a - 12
15.
5c2 + 5c - 60
16. m2n - n2 + 3m2p - 3np
17.
x2 - 4y2 + 12yz - 9z2
18. w2 - 9x2 - 24xy - 16y2
19.
9r4 + 30r2s2 + 25s4
20. u4 - 10u2v2 + 25v4
21.
4p4 + 16p2q2 + 16q4
Rational Expressions
Numerical fractions of the type 1/2, 3/4, 2/3 and 5/7 are common in arithmetic and
are used constantly in everyday living. Algebraic fractions are equally important in
mathematics and in all fields to which algebra is applied. Skill in operations that involve
fraction is essential for progress in any of these fields.
A rational expression, which is another term for fraction, is an expression of the
type a/b where b ≠ 0. The following are examples of rational expressions:
4a
---7
3x2
------2xy
c2 - d2
----------c 2 + d2
(x + y)(x4 + xy + y4)
--------------------------(x + y)(x3 + x2y + y2)
29
In dealing with fractions, it is worthwhile to recall the laws on exponents, operations
involved in special products and factoring as well as to familiarize the following theorem:
a
c
If --- = ---, then ad = bc
b
d
-a
a
---- = - --b
b
(both b and d ≠ 0)
a
a
---- = - ---b
b
(b ≠ 0)
Also, the fundamental principle on fractions should be remembered. It says that "if
the numerator and denominator of a fraction are multiplied or divided by the same
nonzero number, the resulting new fraction is equal to the original fraction". This simply
means that the value of a fraction is not altered or changed if both the numerator and
denominator are multiplied or divided by the same nonzero number or expression. For
example:
a
a(n)
--- = -----b
b(n)
3
3(5)
15
--- = ------ = ---4
4(5)
20
a
a m
--- = -----b
b m
24
24 12
2
---- = --------- = --36
36 12
3
There are three signs associated with any fraction. It is the sign preceding the
entire fraction, the sign preceding the numerator and the sign preceding the
denominator. If two of these signs in a given fraction are changed, the resulting fraction
is equal to the original given fraction. Thus,
a
-a
-a
a
--- = ----- = - ----- = - ---b
-b
b
-b
-x
-(-x)
x
--------- = ---------- = ------(y - x)
-(y - x)
x-y
y3 - x3
-(y3 - x3)
x3 - y3
--------- = - ------------ = --------x-y
x-y
x-y
x2 - y2
x2 - y2
x2 - y2
--------- = - ---------- = - --------y-x
-(y - x)
x-y
30
1.6.1 Simplification of Fractions
A fraction is said to be in its lowest terms if both the numerator and denominator
have no common factors except 1. Such a fraction is called a reduced fraction. In the
above example 24/36 = 2/3, 2/3 is the reduced form of 25/36 since 2 and 3 have no
common factors other than 1.
In reducing a given fraction or a rational expression into its lowest terms, divide
both the numerator and denominator by the product of the factors that are common to
both and obtain the numerator and denominator, respectively, of the reduced fraction. If
the common factors are not clearly visible, it is advisable to factor both the numerator
and denominator first before reducing the given fraction.
Examples:
Reduce the following rational expressions to lowest terms:
1.
65a 4 b 2 5(7)(a 3 )(a )( b 2 ) 5a
6b
42a 3 b 3
6(7)(a 3 )( b 2 )b
2.
x 3 x 2 6x
x ( x 2 x 6) x ( x 3)( x 2) x 3
x 3 3x 2 2x x ( x 2 3x 2) x ( x 1)( x 2) x 1
3.
a 5 a 4 c ab 4 b 4 c
(a 5 a 4 c) (ab 4 b 4 c)
a 4 (a c) b 4 (a c)
a 4 a 3 c a 2 b 2 ab 2 c (a 4 a 3 c) (a 2 b 2 ab 2 c) a 3 (a c) ab 2 (a c)
=
(a c)[a 4 b 4 ] (a 2 b 2 )(a 2 b 2 )
(a c)[a 3 ab 2 ]
a (a 2 b 2 )
a 2 b2
=
a
4.
x 2 7 x 10 ( x 5)( x 2)
x 5
2
2x x 6 (2x 3)( x 2) 2x 3
5.
12x 2 30x 72 6(2x 2 5x 12)
3(2)( 2x 3)( x 4)
2
2
52x 8x 60 4(13x 2x 15) 2(2)( 2x 3)( x 5)
=
3(2x 3)( x 4)
3(2x 3)( x 4)
2[( )( 2x 3)( )(5 x )] 2(2x 3)(5 x )
=
3( x 4)
2(5 x )
31
1.6.2 Addition and Subtraction of Fractions
The sum of two or more fractions having the same denominator is a fraction with
the same common denominator as its denominator and the sum of the numerators as
its numerator. Thus,
a b c a bc
d d d
d
Examples:
1.
2.
2a
6b
a 2b 2a 6b a 2b 3a 4b
ab ab ab
ab
ab
2x 2
3x
5
2x 2 3x 5
x4 x4 x4
x4
To find the algebraic sum of two or more fractions with different denominators, the
least common denominator must have to be determined first. The least common
denominator (LCD) is the least common multiple of all the denominators. To find the
least common multiple of the denominators or LCD, each denominator is written in its
prime factored form, then get the product of all the individual prime factors, each factor
being used the greatest number of times it appears as a factor of some denominators.
Example 1. Find the least common multiple of 6, 8, 18, 27.
Solution: We write each number in prime factor form and observe that
We note that 2 and 3 are the only numbers that occurs those factors and that each
appears in any one number at most three times. Hence LCM is
Example 2. Find the LCM of x2 - x - 2; x2 + x - 6
Solution: Write each expression in prime factor form.
x2 - x - 2 = (x - 2)(x + 1)
x2 + x - 6 = (x - 2)(x + 3)
32
So the LCM is
(x - 2)(x + 1)(x + 3)
The process of finding the algebraic sum of fractions having different denominators
consists of the following steps.
1. Factor the given denominators into prime factors.
2. Determine the least common denominator (LCD). The LCD becomes the
denominator of the resulting fraction.
3. Divide the LCD by the denominator of the first fraction and multiply the
quotient to the numerator of the first fraction. The product becomes the
first term of the numerator of the resulting fraction.
4. Repeat step 3 for the rest of the given fractions.
5. Simplify and determine the algebraic sum of the terms in the numerator of
the resulting fraction.
Hence,
a c ad bc
b d
bd
Examples.
1.
4
x 3 2x 1
4
x 3
2x 1
2
x 2 x 4 x 2 x 2 ( x 2)( x 2) x 2
=
4( x 2) ( x 3) (2x 1)( x 2)
( x 2)( x 2)
=
(4x 8) ( x 3) (2x 2 5x 2)
(x 2)( x 2)
=
4x 8 x 3 2x 2 5x 2
( x 2)( x 2)
=
2x 2 10x 3
x2 4
33
2.
x 2 2xy
y
x
x 2 2xy
y
x
2
2
3( x y ) 6( x y) 4( x y) 3( x y)( x y) 6( x y) 4( x y)
4( x 2 2xy ) 2( x y)( y) 3( x y) x
=
12( x y)( x y)
3.
=
(4x 2 8xy ) (2xy 2 y 2 ) (3x 2 3xy )
12( x y)( x y)
=
4x 2 8xy 2xy 2 y 2 3x 2 3xy
12( x y)( x y)
=
x 2 3xy 2y 2
( x 2 y)( x y)
12( x y)( x y) 12( x y)( x y)
=
x 2y
12( x y)
3x y
x 3y
1
( x 2 y)(3x y) ( x y)( x 3y) ( x y)( x y)
( x y)( x y) ( x y)( x 2 y) x 2 y
( x y)( x y)( x 2 y)
=
3x 2 5xy 2y 2 (x 2 2xy 3y 2 ) ( x 2 y 2 )
(x y)( x y)( x 2y)
=
3x 2 5xy 2 y 2 x 2 2xy 3y 2 x 2 y 2
( x y)( x y)( x 2 y)
=
x 2 3xy 2 y 2
( x 2 y)( x y)
( x y)( x y)( x 2y) ( x y)( x y)( x 2 y)
=
1
xy
34
1.6.3 Multiplication of Fractions
The product of two or more fractions is a fraction whose numerator is equal to the
product of the given numerators and whose denominator is the product of all the given
denominators.
a c e ace
b d f bdf
If possible, factor all the members of the given fractions before the product is
formed. The factors that are common both in the numerator and denominator can just
be cancelled out easily.
Examples:
1.
x 2 x y x 2 xy y 2
x 2 ( x y)( x 2 xy y 2 )
y 3 x y x 2 3xy y 2 y( x y)( x 2 3xy y 2 )
2.
a 2 4b 2
6a 3b a 2 4ab 3b 2 (a 2b)(a 2b) 3(2a b) (a b)(a 3b)
(2a b)(a 3b) 2(a 2b) (a b)(a 2b)
2a 2 7ab 3b 2 2a 4b a 2 ab 2b 2
3.
4.
=
3(a 2b)(a 2b)( 2a b)(a b)(a 3b)
2(2a b)(a 3b)(a 2b)(a b)(a 2b)
=
3(a b)
2(a b)
x 2 3x 2 2x 2 5x 3 3x 2 6x (x 1)( x 2) (2x 1)( x 3) 3x( x 2)
2x 4
(2x 1)( x 2) (x 1)( x 1)
2(x 2)
2x 2 3x 2
x 2 1
=
3x ( x 1)( x 2)( 2x 1)( x 3)( x 2)
2(2x 1)( x 2)( x 1)( x 1)( x 2)
=
3x ( x 3)
2( x 1)
a 2 b a 2b
ab
a 2b
a 2b
ab
2
2
2
2
ab
(a b)(a b) (a b)
(a 2b)(a 2b)
a b
a 4b
=
(a 2b)(a 2b)(a b)
(a b)(a b)(a b)(a 2b)(a 2b)
1
(a b)(a b)
35
=
5.
1
(a b) 2
x 4 4x 8
x4
4( x 2)
4( x 4)( x 2)
2
2x 8 x 16 2( x 4) ( x 4)( x 4) 2( x 4)( x 4)( x 4)
=
2( x 2)
( x 4) 2
1.6.4 Division of Fractions
In order to obtain the quotient of two fractions, obtained the reciprocal of the divisor
and multiply to the dividend.
Hence,
a c d a
b d c b
Examples:
x 2 3x 2
x2 x 2
2x 2 3x 2 x 2 3x 2
1.
2
2x 2 7 x 3 2x 2 3x 2
x x 2 2x 2 7 x 3
2.
=
(2x 2 3x 2)( x 2 3x 2) (2x 1)( x 2)( x 2)( x 1)
( x 2 x 2)( 2x 2 7 x 3) ( x 2)( x 1)( 2x 1)( x 3)
=
( x 2)( x 1)
( x 1)( x 3)
3x 15 12x 18 4x 12 3x 15 4( x 3) 3( x 5)
x3
4x 12 12x 18 x 3
6(2x 3) ( x 3)
=
12( x 3)( x 5)
6(2x 3)( x 3)
=
2( x 5)
(2 x 3)
36
1.6.5 Complex Fractions
A complex fraction is a fraction in which one or more of the terms of one or both
members is a fraction. The following are examples of complex fractions:
3
------2
--3
x
1 + ---y
--------------x+y
4x
2y
------- + ------x+y
x-y
---------------------------x 2 + y2
3 - ------------x2 - y2
A complex fraction can be simplified by reducing both the numerator and
denominator first to single fractions before dividing the numerator by the denominator.
Examples:
1.
1
a 2 1
a
a2
a 2 1
a a
1
a
a3 1
a3 1
a 2
2
a
a
a 2 (a 1)(a 1)
a (a 1)(a 2 a 1)
=
a (a 1)
a2 a 1
y2 x 2
x
(y 2 x 2 )
y
y
y2
y2 x 2
(y 2 x 2 )
x
x
x
y
2.
=
x2
y
=
x
y
37
Exercises. Perform the indicated operation on the following algebraic expression.
A.
B.
C.
Simplify the following fractions to lowest terms.
1.
8x 4 y 7
12 x 6 y 3
2.
25a 4 x 3 y
225a 2 xy 3
3.
a 2 b2
a 3 b3
4.
2x 2 7 x 15
x 2 3x 10
5.
(r 3s)( r 2 6rs 8s 2 )
(r 2s)( r 2 8rs 15s 2 )
6.
pr 2qr 2ps 4qs
2pr 4qr ps 2qs
Determine the algebraic sum of the following rational expressions and reduce to
lowest terms.
4s
5r 9r 2 8s 2
21r 14s
42rs
7.
c
a
b
6ab 4bc 3ac
9.
3
2
4
(m 2n )( m n ) (m 2n )( m n ) (m 2n )( m 2n )
10.
2a b
a 4b
a 7b
2
2
2
2
a 3ab 2b
a 4ab 3b
a 5ab 6b 2
8.
2
Multiply or divide the following rational expressions and then simplify.
11.
15x 3 z 4 12x 2 y 5 12x 4 yz
4 yz
4y 2
5z 2
12.
w 3x 2 w 4 y2 w 6z2
y4 z0 x 3z5
xy 3
13.
9x 2 y 2
(9x 2 3xy )
x 3y
14.
(p 2 5p 14)
15.
3p q
p 2 qr
2qr 6r 2
pq 3pr p 2q q 3 9qr 2
p2 4
5p
38
16.
D.
c 2 cd 6d 2 c 2 cd 12d 2 c 2d
cd
c 2 cd 12d 2 c 2 cd 2d 2
Simplify the following complex fractions:
1.7
1
1
xy xy
xy xy
17.
x 2 xy y 2
1
x 2 y2
1
18.
1
x 1
1
1
1
x 1
Radical Expressions
The customary notation for the principal rth root of the number a is r
is called a radical of order r. The number a is the radicand and r is the index of the
radical.
If a is positive (a > 0), the principal rth root of a is the positive real rth root. If a is
negative (a < 0) and r is an odd integer, the principal r th root of a is the negative real rth
root. If a is negative (a < 0) and r is an even integer there is no real r th root of a since it
is an imaginary number already.
For example:
42
3
82
3
243 3
Therefore, a1/r is defined as the principal rth root of a. Hence, a1/r = r
Generally,
a8/ t r a8
1.
a
8
r
In many cases, it is more advantageous to express a quantity in terms of radicals
than in terms of rational exponents. The laws of radicals follow directly from the
previous definition and theorems on exponents. If m and n are positive integers and a
and b are also positive:
(n a ) n a
n
ab (ab)1 / n a 1 / n b1 / n n a
(1)
a
b
(2)
39
n
a
a/b
b
n n
1/ n
a a1/ n
1/ n
a1/ n
1/ n
b
n
a
n
b
(3)
a 1 / an an a 2
(4)
By the use of the above guides, exponential expressions can be transformed to
radical expressions and vice-versa.
Examples:
1.
91/2 =
8 =3
2.
27 2 / 3
27
3.
(128) 5 / 7 (7 128 ) 5 (2) 5 32
4. 813 / 4
2
3
(3) 2 9
1
1
1
1
4
3
3/ 4
27
81
( 81) (3)
5.
x 1 / 4 ( x 2 / 3 ) x (1 / 4 2 / 3) x 11/ 12 12 x 11
6.
x 1 / 4 x 2 / 3 x (1 / 42 / 3) x 5 / 12
7.
4 2 / 3 (4 2 )1 / 3 161 / 3
= (8 2)1 / 3 81 / 3 (21 / 3 )
=
3
8
3
2
=2 2
8.
2a 1 / 3 b 2 / 3 2(ab 2 )1 / 3 2 3 ab 2
9.
a 2
3a 2 / 5
3 3
b3 / 5
b
1/ 5
a2
3 3
b
5
1
x
5 / 12
1
12
x5
40
1.7.1 Simplification of Radical Expressions
A complete simplification of radicals by the use of the laws on radicals will yield:
1. no factors which are perfect nth powers under a radical whose index is n
2. no fraction under a radical sign or no radical appears in the denominator
that is the denominator is rationalized.
3. the smallest possible index of the radical
Any radical that satisfies the above conditions is said to be in its simplest form.
Examples:
8 4(2) 4
1.
2.
4
5.
32 4 16(2) 4 16
4
2 24 2
192a 3 b 7 64(3)a 2 a b 6 b 64a 2 b 6
3.
4.
2 2 2
3
81x 5 y 7 3 27(3) x 3 x 2 y 6 y 3 27 x 3 y 6
(a 2 b 2 b 2 c 2 ) b 2 (a 2 c 2 ) b 2
3
3ab 8ab 3 3ab
x 2 y 3xy 2
3
3x 2 y
(a 2 c 2 ) b (a 2 c 2 )
1.7.2 Rationalizing Radical Expressions
Using the laws on radicals, it is always possible to convert a fraction whose
denominator is a monomial or a binomial that contains a radical expression to an equal
fraction in which the denominator contains no radical expression. This process is called
rationalizing the denominator. In many computations involving radical expressions, it is
advisable to rationalize all denominators as a first step.
An expression which when multiplied to a radical expression yields an expression
free of radical sign is called rationalizing factor or conjugate.
41
Example A: Expression
Rationalizing factor or conjugate
1.
3
3
2.
8
2
3.
3
3
3
9
4.
3
5
3
25
5.
3
4
3
2
a b
6.
7.
3
a b
2 3
3
2 3
8. 5 2
5 2
9. 5 2 6 3
5 2 6 3
10. 2 a 3 b
2 a 3 b
Example B: (Monomial Denominators)
Note: The use of least Rationalizing factor should be observed to avoid yielding an
expression not in the lowest term.
2
1.
3
x
y
2.
3.
3
4.
4
2
3
x
y
3a
2b 2 c
1
2 2
b c
3
3
3
3
2 3
x
y
3a
2b 2 c
4
4
1
b2c2
3
y
y
3
3
2 3
3
( y)
4abc 2
4bc 2
1
4
x ( y)
b2c2
2
3
3
xy
y
12abc 2
8b 2 c 2
4
b2c2
4
b2c2
3
12abc 2
2bc
4
b2c2
4
b4c4
4
b2c2
bc
42
5.
x 2 y2
2 x ( x y)
x 2 y2
2 x x y)
xy
( x y)
( x 2 y 2 ) ( x y)
2x[ ( x y) 2 ]
( x y)( x y) ( x y) ( x y) ( x y)
2 x ( x y)
2x
=
Example C: (Binomial Denominators)
1.
2.
4
5 1
4
5 1
a b 2 ab
a b
5 1
5 1
4( 5 1) 4( 5 1)
5 1
5 1
4
a b 2 ab
a b
a b
a b
(a b 2 ab )( a b )
( a b )( a b
=
a a b a 2 a 2 b a b b b 2 ab 2
ab
=
a a b a 2a b a b b b 2b a
ab
=
a a b a 2b a 2a b a b b b
ab
=
a (a b 2b) b (2a a b)
ab
=
a (a b) b (a b)
ab
=
=
(a b)[ a b ]
ab
a b
43
1.7.3 Multiplication and Division of Radical Expressions.
Obtaining the product and quotient of radical expressions is governed by the
second and third law on radicals. If the radicals have the same index, their radicans
can be multiplied and placed under a single radical sign. However, if the radicals are of
different indexes, they must be converted first to radicals having similar indices before
multiplication of the radicands can take place. This is always possible with the aid of
equivalent expressions having rational exponents.
Division of radicals is handled in a similar manner. Again, the radicals must have
similar indices first before division of the radicals can be carried out.
Examples: (Multiplication)
1.
14 21 14(21) 7(2)(7)(3) 7 2 (6) 7 6
2.
15ax 3 ( 45a 2 xy 3 ) 152(15)(3)a 2 ax 4 y 2 y (15) 2 a 2 x 4 y 2
3.
( x 2 y 2 ) ( x y) ( x y)( x y)( x y) ( x y)( x y) 2 ( x y) ( x y)
4.
( 5 2 3 )( 5 3 3 ) ( 5 ) 2 3
6x 3
5.
3
5 6( 3 ) 2 5 15 6(3) 13 15
4 x 4 y 2 (6 x 3 ) 1 / 2 ( 4 x 4 y 2 ) 1 / 3 (6 x 3 ) 3 / 6 ( 4 x 4 y 2 ) 2 / 6 (6 3 x 9 ) 1 / 6 ( 4 2 x 8 y 4 ) 1 / 6
= (2333x9)1/6(2222x8y4)1/6 = (26 2 33x12x5y4)1/6 =
= 2x 2
6
54 x 5 y 4
Examples: (Division)
1.
18 12
6
4
2.
4
24a 3 b
8ab 3
3ay 15ax 2 y 3ay
18(12)
18(2) 36 6
6
4
24a 3 b 4 3a 2 4 3a 2 4 3a 2
4
4
8ab 3
b2
b2
b2
4
=
3a 2 b 2
b
4
3a 2 b 2
4
b4
6
2 6 x 12
6
2 33 x 5 y 4
44
3.
635
2 3
635
2 3
3
3
635
3
2( 3 ) 2
635 3 3
5 3
2(3)
= (5)1 / 3 (3)1 / 2 (5) 2 / 6 (3) 3 / 6 (5 2 )1 / 6 (33 )1 / 6
=
6
4.
12
3
3
5 2 3 2 6 25(27) 6 675
(12)1 / 6
(12)1 / 6
(12)1 / 6
2 2 (3)
6
1/ 2 1/ 3
(3) 3 / 6 (2) 2 / 6 (33 )1 / 6 (2 2 )1 / 6
2 2 (3) 3
2 (3) (2)
=
5.
6
6
1
(3) 2
( x y ) ( x y)
(x 2 y 2 )
=
=
6
3
1
(3) 2
1
3
3
( x y) ( x y)
(x 2 y 2 )
3
3
9
9
3
9
3
2
3
9
3
(x 2 y 2 )
(x 2 y 2 )
( x y) ( x y)( x 2 y 2 )
(x
2
y2 )
2
( x y) ( x y)( x y)( x y)
(x 2 y 2 )
=
( x y) ( x y) 2 ( x y)
( x y)( x y)
=
( x y)( x y) ( x y)
( x y)( x y)
=
( x y) ( x y)
xy
1.7.4 Addition and Subtraction of Radical Expressions
In adding or subtracting radical expressions, all similar radicals (that is, those
containing the same index and radicand) are combined into single terms.
If the radical expressions to be added or subtracted have different orders or
radicands, the following procedure is suggested.
45
1.
2.
3.
Simplify each radical expression and rationalize all denominators.
If it is possible to convert one or more radical expressions to a lower order, do it.
If possible, assemble the resulting radical expressions into one or more groups
composed of radical expressions of the same or radicands then apply the
distributive axiom to each group.
Examples:
1. 5 a 3 a 2 a a (5 3 3) 6 a
2.
108 48 3 36(3) 16(3) 3 6 3 4 3 3
3 (6 4 1)
=
=9 3
3. 4 12 5 8 50 7 48 4 4(3) 5 4(2) 25(2) 7 16(3)
= 4(2) 3 5(2) 2 5 2 7(4) 3
= 8 3 10 2 5 2 28 3
= 8 3 28 3 10 2 5 2
3 (8 28)
=
2 (10 5)
= 20 3 5 2
= 5 2 20 3
= 5( 2 4 3 )
4. 8
1 3
3 2
108 4 9
=
8 1
3
8
3
=
=
=
3
36(3)
2
3
3
9
3
8 3 18
(6) 3 3
3 3
2
( 3) 2 2
8 3
8
9 3 3 3 9 1
3
3
8
8 24
3 8 3
3
3
32 3
3
46
5.
8a 3 b 3 +
3
2
2
ab
3 8a 4 b 4 4 4a 2 b 2 4(2)a 2 ab 2 b 3 ab
ab
ab ab
ab
3 8a 3 a b 3 b
= 2ab 2ab 3 ab
=
3
=
3
4a 2 b 2
2ab
2ab 3 ab 2ab
ab
ab 2ab 3 ab 2ab
2ab
2ab
2ab
ab
1
ab (a 2ab) 2ab 2ab
1
ab
Exercises. Perform the indicated operation. Rationalize the denominator if possible.
A.
B.
Convert the following exponential expressions to radical expressions.
1.
x4/5
3.
s3/4t1/4
m2/3
3.
b 3 / 5
d 2 / 5
Simplify the following radical expressions.
5.
7.
C.
2.
112d11f 14
6.
169x 6 z 2
y4
8.
4
3
81z 2 x 6 y 5
9
x 6
x
Rationalize the following radical expressions.
9.
11.
3
7u 2
4v 2 w 4
x
x y
10.
12.
r 2 s2
2s r s
2 7 3
3 7 5 3
47
D.
Determine the product and quotient of the following radical expressions.
6x 4 y
13.
15.
E.
3
9x
6
3xy 2
27 x
14.
4
3
6p 3 r
3
4p 2 r 5
(x 2 y 2 )
16.
( x y) ( x y)
Find the algebraic sum of the following radical expressions.
17.
2 32 18 8
19.
4(x y) 2 9(x y) 3 (x y)
20.
3
a 4b
1
3
2
a b
2
3 3 ab 4
18.
3
375 72 3 192 98
48
Chapter 2
2.1
RELATION AND FUNCTIONS
Relation
Relation is the dependence, association, connection or interrelationship between
variables, components, conditions or situations.
Some statements which shows relation:
1.
2.
3.
4.
Energy needs of a person depends on the temperature of the day.
The price of goods rise when the supply is low.
Living expenses increases with each additional child on the family.
The depth of irrigation water affected weed population
2.1.1 Picturing Relations:
a.
By tables
P points
S (segments)
2
1
3
3
4
6
5
10
6
7
8
The table shows the relations between the number of segment for a given number
of points.
b.
By ordered pairs
Let S = {a, b}, T = {1, 2, 3}
S x T, read S dot T, S cross T or the cartesian product of S and
T = {(a, 1)(a, 2)(a, 3)(b, 1)(b, 2)(b, 3)}
T x S = {(1, a)(1, b)(2, a)(2, b)(3, a)(3, b)}
S x S = {(a, a)(a, b)(b, a)(b, b)}
T x T = {(1, 1)(1, 2)(1, 3)(2, 2)(2, 1)(2, 3)(3, 3)(3, 1)(3, 2)}
(a, 1)(a, b) are called ordered pairs.
The first element/component is the input.
The second element/component is the output.
Consider the set {0, 1, 2, 3, 4, 5, 6, 7, 8}
List the pair of relation satisfying the condition.
a) The first component is greater than the second component.
{(2, 1)(3, 1)(4, 1)(5, 1)(6, 1)(7, 1)(8, 1)...}
49
c.
By Mathematical Sentence or Equation
Given the set of ordered pairs {(1, -1)(-1, 1)(2, -2)(3, -3)(-3, 3)}
Write the sentence or equation that describe the relation.
The second number is the additive inverse of the first number. In equation
form, x + y = 0.
d.
By arrow diagram
x
y
a
3
4
5
6
14
13
12
11
-1
b
1
1
-2
4
2
-3
9
3
e.
By Graph
Exercises:
1.
Consider the set {0, 1, 2, 3, 4, 6, 8}. List the ordered pairs of the relations satisfying
the conditions:
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
The first component is less than the second component.
The first component is a multiple of the second component.
The first component is the reciprocal of the second component.
The first component is identical with the second component.
The first number is twice the second number.
y is the square of x.
x is the square of y.
x+y 7
2x y
x+y 7
50
2.
Write the sentence that describes the relation:
1.
2.
3.
{(1, -1)(-1, 1), (2, -2), (3, -3), (-3, 3)}
{(25, 5), (16, 4), (9, 3), (3, 2), (1, 1)}
{(0, 0), (1, 5), (2, 10), (3, 15), (4, 20)}
4.
x
y
1
5
2
6
3
7
4
8
x
y
1
9
2
8
3
7
4
6
x
y
-8
-4
-4
-2
6
3
10
5
5.
6.
2.2
26
13
Function
Function - is a special kind of relation. It is a rule that assigns to its input number
exactly one output number. The set of all input numbers to which the rule applies is
called the domain of the function. The set of all output number is called the range
of the function.
Example 1. y = x + 2 defines y as a function of x. X is the input or independent
variable or domain. Y is the output or dependent variable or range.
Example 2. x =
y
defines x as a function of y.
2
Letters are used (capital or lower case) to represent function rules.
Input
f(x) reads “f of x” means the output number in the
Range of f that corresponds to the input number x
in the domain.
f(x)
Output
Thus the output number corresponds to y. But since y = x + 2 in example 1 we
may write y = f(x) = x + 2 or simply f(x) = x + 2.
Other sources defines function as the set of ordered pairs such that each element x
of the domain is paired with one and only one element y of the range and no two
ordered pairs have the same first element but different second element.
51
Examples:
a) {(b, b), (a, o), (d, y)} - function
b) {(n, g), (i, i), (c, r), (e, 1)} - function
c) {(1, 2), (3, 2), (4, 3)} - function
d) {(1, 2), (1, 3), (2, 3)} - not a function
Because the first 2 ordered pairs have the same elements.
Domain - is the set of all the first elements in a set of ordered pairs.
Range - is the set of all the second elements in a set of ordered pairs.
In example a, the domain is (b, a, d) and the range is (b, o, y).
In mathematics, the rule relating the elements of the domain to elements of the range is
usually given as an algebraic expression. The domain can be the set of real numbers
or any subset of the set of real numbers and is sometimes limited by the algebraic
expression of the rule. For instance,
y
1
x
has a domain of all real numbers except zero, i.e. (-4, 0) U (0, + -4).
Also,
y x
has a domain of all non-negative real numbers, i.e. (0, -4). In other words, from the
replacement set for x (domain), we exclude numbers which would lead to division by
zero or square root of negative number which is an imaginary number. To determine
whether a relation, defined by an equation, is a function or not, solve the equation for
the variable using values for x from the set of real numbers. If for each x a unique y is
paired or for many x's there are many y values, then the relation is a function. On the
contrary, if one is paired with many y's, the relation is not a function.
When a relation is in graphical form, it can be determined whether its a function or
not by "vertical line test". If the vertical line drawn on any part of the graph touches at
only one point, then the graph is a function. If the vertical line touches at more than one
point on the graph, the graph is not a function.
Examples:
vertical line touches at only one point A.
hence, a function
52
vertical line touches at 2 points on the
graph, hence, not a function.
Functional notation:
Function are denoted by small letters like f,g,h, etc. The notation f(x) is read "f of x"
or "f at x" and means "the element in the range of f associated with the elements x in the
domain" (image of x).
For Example,
y = 2x is also, a function
f(x) = 2x which means "the image of x(y) is a number that is twice x"
Any other letter can be used in place of "f", say "g" to denote a function, g(x) = 2x.
Evaluating a function:
To evaluate a function, simply replace x by its value everywhere in the equation
then do the indicated operation.
Example 1. Let f(x) = 2x - 2 then
f(3) = 2(3) - 2 = 4
f(2t) = 2(2t) - 2 = 4t - 2
f(3 + a) = 2(3 + a) - 2 = 6 - 2a - 2 = 4 - 2a
2. Express the area of a circle as function of its circumference C.
Solution: C = 2r
A = r2
From: C = 2r,
C
A =
2
=
C 2
4 2
=
C2
4
r = C/2, 50
2
53
Exercise 1.
In each of the following relations consider (from the set of real number) the
replacement set for x as the domain and the replacement set for y as the
range. Find the domain, range, and tell whether each is a function or not.
1. y = x2 – 2
2.
y 9 x2
3.
y
4.
x2
y
x
5.
y
6.
1
x
y
7. x2 + y2 = 9
x2
x 3
8. y
1
x4
y
1
9.
x2
x 2
10. x y 2
x
Exercise:
1.
Given the relation, y = x2 - 2x + 2, find:
a) f(2)
b) f(-3)
c.
f(0)
2.
If f(x) = 2x2 - 2x + 2, find the value of f(x) when, x = -3, 0: 3: 5.
3.
If f(x, y) = y(x2 - 2y + 3), find:
4.
If f(x, y) = 4 +
5.
Express the perimeter P, of a rectangle as a function of its sides, a and b.
Compute the perimeter when a = 2 and b = 4.
6.
Express the area of a square as a function of its perimeter P.
7.
Express the circumference, C, of a circle as a function of the radius, r.
Compute the value of C when r is 3 ft. What is the radius of a circle when C =
62 ft. ( = 3.14).
8.
Express the area, A, of a circle as a function of its radius, r. Find A(3),
A(15/2).
a) f(-3, -5)
b) f(4, 2)
x 2y
, find: f(-2, -3)
y2
54
9.
The perimeter of a rectangle is 24 in. Find L as a function of its width.
10. Which of the following are relations/functions?
a. {(1, 5), (5, 1), (3, 4), (4, 3)}
b. {(-0, 1), (-1, 0), (0, 1), (1, 2)}
c. {(1, 1), (1, -1), (4, 2), (4, -2), (9, 3), (9, -3)}
d. {(x, y)/y = 3x - 1}
e. {(1, 4), (2, 5), (3, 6)}
f.
g.
i.
j.
h.
11. Give the equation that describe the functional relationship.
1. {(1, 3), (3, 9), (5, 15), (-7, -21)}
2. {(2, -2), (5, 1), (7, 3), (9, 5)}
3. {(1, 1), (-2, 2)(-3, 3)....}
4. The circumference of a circle is a function of a radius
5. The area of a square is a function of its side.
6. The interest earned is a function of the amount of deposit.
12. Given:
a. f(x) = 3x = 2
b. f(x) =
x2
x 1
find f(1/2), f(2), f( 3 )
find f
3
, f(2 3 ), f(x + 2)
4
c. f(x) = x2 - x + 1 find f(x + 1), f(x - 1/3), f(x - 3)
d. H(x) = x1/3
find H(0), H(8), H(64), H(5)
e. g(x) = 2x
find g(0), g(-1), g(2), g(-2)
55
Chapter 3
LINEAR EQUATIONS
An equation is an algebraic sentence containing two expressions called members
(the left and right members) separated by an equal sign (=). It denotes equality of two
expressions.
Ex: 8x - 7 = 3 + 3x
(left member) (right member)
An equation is called an identity if its two members have same values.
Ex: 1 + 2 = 3 }
x + x = 2x} are identities
(This is an equation which is satisfied by all the values of the variables for which
both members of the equation are defined).
An equation containing literals is called a conditional equation if its two members do
not have the same value.
(This is an equation which is satisfied by some, but not all of the values of the
variables for which the members of the equations are defined).
Ex: x - 3 = 4
x2 - 2x - 15 = 0
Root/Solution set - the set of all elements which when substituted to the variables
will make the equation true.
3.1
Solution of Equations
The process of finding the root or roots of an equation is known as solving the
equation. This process makes use of one or combinations of the different axioms for
equality.
LINEAR EQUATION - an equation of the first degree in its variable or variables.
It is linear if its members are expressions of the following types.
1.
A polynomial is linear if none of its terms contains more than one literal factor.
Ex. 2x + 3y - 4z
2.
A reduced fraction is linear if its denominator is a nonzero integer and its numerator
is linear
56
Ex.
3.
A sum of linear expression is linear
Ex.
4.
2 x 3y
x 1
or
5
2
x 5y
2 x 3y 6 x 2 y
and
2 3
4
5
Any expressions that has the same value as a linear expression is linear
Ex. 2(3x - 5) and 5(6x - y) - 7x + 8
Warning: A product or quotient of linear expressions is not necessarily linear.
3.2
LINEAR EQUATION IN ONE UNKNOWN:
- an equation which involves one literal and its in the first degree.
- Such that linear equation in variable x is an equation in x which can be expressed
or reduced to the form ax + b = 0; a 0.
STEPS TO FOLLOW IN SOLVING A LINEAR EQUATION IN ONE UNKNOWN
1.
If the equation is fractional, clear the equation of fractions by multiplying both sides
of the equation by the Least Common Multiple of all the denominators (LCM).
2.
Simplify both sides of the equation by carrying out indicated operations of
multiplication if there are any and by collecting like terms.
3.
Put all the terms containing the unknown to one side of the equation and all the
other terms of the other side by transposition.
TRANSPOSITION -
process of transferring a term from one side of the equation to
the other side by changing its sign.
4.
Divide both sides of the equations by the coefficients of the unknown.
5.
Check the solution by substituting the value obtained in the original equation.
Reject all extraneous roots.
FOR EQUATIONS INVOLVING RADICALS
To solve here are the steps:
1.
Put one of the radicals (if there are more than one) on one side of the equation and
all the other terms on the other side by transposition.
57
2.
Square both sides and simplify
3.
Repeat steps 1 and 2 (if necessary) until the resulting equation has been cleared of
radicals.
4.
Solve the resulting equations.
5.
Check by substituting the value obtained to the original equation and discard
extraneous roots.
Tell whether the equation is linear or not if linear solve for the roots
1.
3x 5 x 3
6
5
4
2.
0.4x
2.2
0.2 x
9
3
3.
0.05(x + 2) = 0.03(2x – 4)
4.
2
1
16
2
x 5 x 5 x 25
5.
3
2y
5
2
y2 4y
y2
6.
3
7.
x 5 2
3y 1 2y 6
8. 2 t 1 3t 1
9.
2x 3 1 x 2
10.
x 5 2 2x 5
11.
2x 3 1 2x 3
12.
x 3 x 5 4
58
Verbal Problems:
1.
Narda and Rosie went to their backyard to pick flowers together they picked 60
flowers. Rosie picked 12 more than half the number of flowers picked by Narda.
How many flowers did each girl picked?
2.
Joey spent 180 buying a pair of Levis pants for each of his two sons. If one pair
cost 45 more than the other, how much did each cost?
3.
Aiko saves 3.75 in 25 centavo pieces and 10 centavo pieces. If she has 18 coins in
all, how many coins of each kinds are there?
4.
Two rice fields have the same area. One farmed by scientific method yield 60
canvas of palay more than twice the produce under the ordinary method. If the
total harvest was 120 cavans, what was the harvest of each field?
5.
If a liter of cooking oil is 90% pure how many liters of coco oil should be added to
make a solution of 60% cooking oil?
6.
A farmer can plow a field in 2 days using a tractor while his brother can make it in 8
days using a plow. If they work together, how many days can they finish the job?
7.
In 15 years Nena will be 3/2 times as old as she is now. What is her present age?
8.
Two cars started from the same place and 90 in opposite direction. One travel 12
km faster than the other. Find the speed of each if they are 324 km apart after 3
hours.
9.
Mr. Jose invested 1/3 of a certain amount at 6% and 1/4 of it at 8%. If he received
480 income how much did he invest at each?
10. If 4 consecutive numbers, the product of the second and fourth exceeds the
product of the first and third by 143. Find the numbers.
3.3
LINEAR EQUATION IN TWO UNKNOWNS
An equation that can be reduced to the form ax + by + c = 0 where a, b, c are
constants, and a and b not zero is a linear equation in x and y. (If a or b is zero the
equation is reduced to linear equation in one unknown).
The solution of a linear equation in two unknowns is not unique since all the
ordered pairs that satisfies the equation is a solution of the equation. Consider the
equation 2x + y = 9, the ordered pairs (1, 7) and (2, 5) are just two of the possible
solutions of the equation. So we may denote the solution set as {(x, y)/2x + y = 9}
59
3.3.1 GRAPH OF LINEAR EQUATION IN TWO UNKNOWNS
As its name suggests the graph of linear equation in two unknown is a straight line.
To graph a straight line we only need to know two distinct points called the x and y
intercepts.
Definition:
Y - intercept graphically, is the value on the y-axis where the line passes.
algebraically, it is the value of y when x = 0.
x – intercept graphically, it is the value on the x-axis, where the line passes.
algebraically, the value of x when y = 0.
Note:
1.
Only non-vertical lines has y-intercepts because vertical lines will not pass thru the
y-axis.
2.
Only non-horizontal line has an x-intercept.
3.3.2 SLOPES OF THE LINE
Slope is one of the fundamentals of a line. It is the inclination of a line with respect
to the x-axis. It is the ratio of the rise and run where rise is the change in Y and run is
the change in X. Letting m = slope, we have
m=
y 2 y1
where (x1, y1) and (x2, y2) are coordinates of two distinct points
x 2 x1
If the slope is positive the line rise to the right and when its negative it implies that
the line rises to the left. The slopes of parallel lines are equal while the slopes
perpendicular lines are negative reciprocals with each other or the product of their
slopes is -1. m1 m2 = 1
3.3.3 FORMS OF LINEAR EQUATION
The standard form of linear equation in two variables is Ax + By + C = 0 where A, B,
C are constants and A and B not both equal to zero.
1.
Slope - intercept form - the line whose equation is in the general form: y = mx + b;
m and b are constants
60
where m = slope
b = y-intercept
y = mx + b is in the slope-intercept form. The standard form can be
written in slope-intercept form by solving the Y in terms of X.
2.
Point-slope form
The equation of the line passing thru (X1, Y1) with slope m is y - y1 = m(x - x1);
where x and y are the coordinates of any point on the line and (x1, y1) are
coordinates of specific point on the line.
3.
Two-point form - when two points are given; (x1, y1) and (x2, y2), the equation is:
Y Y1
4.
Y2 Y1
( X X1 )
X 2 X1
Intercepts form:
x y
1
a b
where: a is the x-intercept
b is the y-intercept
Exercises:
A.
Determine which equation are linear:
1.
3x + 5y + 15 = 0
6.
6y - 4/y = 0
2.
3x - 12 = 0
7.
16 + 4y = 0
3.
2x + 4y2 = 12
8.
y = -6/16x - 0.2
4.
5x - 4xy = 12
9.
y = 15/7x + 2.2
5.
3y/3x = 5y + 2
10. y = 1.5x - 3
61
B.
C.
D.
E.
Determine the x and y intercept. Then draw the graph:
1.
y = 4x + 5
6.
3y = 9
2.
4x = 2y + 16
7.
3x = -15
3.
5y + 25 = -50
8.
4y + 3x = 12 + 3x
4.
3x + 2y = 6
9.
2x = 10
5.
5y - 4x = 20
10. y - 5/2x + 4
For each pair of points find the slope:
1.
(3, 8)(9, -4)
4.
(3.2, -12.8)(3.2, 2.4)
2.
(-8, -7)(-9, -12)
5.
(17, -12)(-9, -15)
3.
(-16.3, 12.4)(-5.2, 8.7)
Which equations have no slope and which have zero slope:
1.
3x = 12 + y
5.
12 - 4x = 9 + x
2.
5x - 6 = 15
6.
2y - 4 = 35 + x
3.
5y = 6
7.
3y + x = 3y + 2
4.
y - 6 = 14
8.
3y - 2x = 5 + 9y - 2x
Tell whether the graph of each pair of equations are parallel or perpendicular:
1.
x+6=4
y - x = -2
---------------
5. x = 6
6-x=8
-------------
2.
2x + 5y = 8
6 + 2x = 5y
-----------------
6. 5y = x - 7
5y = 3 + x
---------------
3.
3x + 2y = 6
3x - 2y + 1 = 0
------------------
7.
4.
y + 3 = 5x
3x - y = -2
---------------
y = 2x - 3
y=x
----------------
62
F.
Find the equation of the line, given:
1.
m = 4, (3, 2)
6.
(1, 4)(5, 6)
2.
m = -2(4, 7)
7.
(-3, -3)(6,6)
3.
m = 0, (0, -7)
8.
(0, 0)(7, 3)
4.
m = -1/2(8, 0)
9.
(-1/4, -1/2)(-2, -1)
5.
m = 3/2(5, 4)
10. (1/6, 0)(0, -3/2)
11. (3, 7) and parallel to x + 2y = 6
12. (-4, -5) and parallel to 2x - y = -3
13. (-6, 2) and parallel to 3x - 9y = 6
14. (2, -1) and parallel to 5x + 7y = 8
15. (2, 5) and perpendicular to 2x + y = -3
16. (-3, -5) and perpendicular to 5x -2y = 4
17. (0, 9) and perpendicular to 2x + 5y = 7
18. (3, -2) and perpendicular to 3x + 4y = 5
19. (-3, -4) and perpendicular to -3x + 6y = 2
63
Chapter 4
INEQUATIONS
An inequality is any sentence having one of the words <, >, # or $. A solution of an
inequality is any number making it a true sentence. The set of all solution is called the
solutions set. When we have found the set of all solutions of an inequality, we say that
we have solved the inequality.
4.1
Addition Principle of an Inequality
If a < b is true, then a + c < b + c is true for any number c. Similar statements hold
for >, and .
4.2
Multiplication Principle for Inequalities
If a < b is true, then ac < bc is true for any positive number c and ac < bc is true for
any negative number c. Similar statements holds for >, and .
*The most important thing to remember is that if you multiply an inequality by a negative
number. You must reverse the inequality symbol.
Examples:
1.
Solve x + 3 > 6 then graph
x + 3 > 6 using addition principle
x + 3 + (-3) > 6 + (-3)
x>3
{x|x > 3}
2.
3y < 3/4 multiply by 1/3
1/3 3y < 3/4 1/3
y < 1/4
{y|y < ¼}
3.
-4x < 4/5 multiply -?
-1/4 -4x < 4/5 -1/4
x > -1/5
{x|x > -1/5}
4.
-5x -80 multiply by -1/5
-1/5 -5x -80 -1/5
x 16
(x|x 16}
64
Using the principle together:
Solve 16 - 7y 10y - 4
16 - 7y 10y - 4
-16 + 16 - 7y -16 + 10y - 4 Adding -16
-7y 10y - 20
-10y - 7y -10y + 10y - 20 Adding -10y
-7y -20
-
1
1
1
(17 y) (20) multiply by
17
17
17
y
20
17
Exercises: Find the solution set.
1.
x+8>3
11. 0.2y + 1 > 2.4y - 10
2.
a + 9 -12
12. 2x - 3 <
3.
y - 8 > -14
4.
x - 11 -2
5.
0.3x < -18
6.
-9x -8.1
7.
8.
5y + 2y -21
9.
0.4x + 5 1.2x - 4
3
5
x
4
8
10. 3x
1 3
2x
8 8
13.
13
x + 10 - 1.25x
4
5
3
y
6
4
14. 0.5x < 25
15. –8y 3.2
65
4.3
Graphs of Inequation
To graph a linear inequality, first replace the inequality sign by an equal sign. Find
the intercepts of the line then graph the resulting line. If the inequation symbol is > or <
draw the line dashed. If the inequality symbol is # or $ draw a solid line. The graph
consist of the region represented by the half plane hence several possible answer could
be taken.
Problem Application in Inequations
1.
In a math course, there are 3 test you must get a total score of 270 or more for an
A. You get scores of 91 and 86 on the first 2 test. What scores on the last test will
give you an A.
Solution:
-------------------------------Test 1 Test 2 Test 3
--------------------------------91
86
x
---------------------------------
--------------------------Total 270 or better
---------------------------
Eq. 91 + 86 + x 270
2.
On your new job as a salesman you can be paid in two ways.
Plan A. A salary of P600 per month plus a commission of 4% of sales.
Plan B. A salary of P500 per month plus a commission on sales of 6% in excess
of P10,000.
For what amount of sales is plan A better than plan B if we assume that sales are
always more than P10,000
Solution:
Plan A
Plan B
P600 salary 4% of sales
P800 salary 6% of sales other P10,000
Total = 600 + 4% of sales
Total = P100 + 6% of sales other P10,000
Let S = total amount of sales for the month.
Income of Plan A > Income of Plan B
Eq. 600 + 4% S > 800 + 6% (S-10,000)
66
3.
A mason can be paid in 2 ways.
Plan A. P300 + 3 pesos per hour
Plan B. straight P8.50 per hour
Suppose a job takes n hours. For what values of n is plan B better for the mason.
Verbal Problems:
4.4
1.
If 3 is added to 9 times any number, the result is always less than 30. What is
the set of numbers.
2.
Eugenio is 5 years older than Lily. The sum of their ges is less than 43. At
most, how old is Lily?
3.
Cita has a certain number of stickers. Twice his number and 8 more will make
her collection at most 60. At most how many stickers had she originally?
4.
A family consume at least one sack of rice in 3 months. If a sack is 50 kg. at
least how many kg. does the family consume per day?
5.
John is 2 kg heavier then Mely. The sum of their weights is less than 40 kg. at
most how heavy is John?
Compound Inequalities
Compound-inequality is made up of two or more inequalities. The sentence -2 < x
< 3 mean -2 < x, x < 3. In order for a number to be a solution it must satisfy both of the
simple inequalities.
Solve -3 < 2x + 5 < 7
-3 < 2x + 5, 2x + 5 < 7
Solving Separately: -3 + (-5) < 2x + 5 + (-5); 2x + 5 + (-5) < 7 + (-5)
-8 < 2x
2x < 2
-4 < x
x<1
Solution set
-4 < x < 1
The shorter method:
-3 < 2x + 5 < 7
-3 + (-5) < 2x + 5 + (-5) < 7 + (-5) Adding (-5)
-8 < 2x < 2 Multiply by ?
-4 < x < 1
67
Exercises:
Graph
Solve:
1.
-7 y -3
1.
-2 < x + 2 < 8
2.
1<x<6
2.
1 < 2y + 5 9
3.
6 > -x - 2
3.
-10 3x - 5 -1
4.
-2 x < 5
4.
2<x+39
5.
-4 - x < 3
5.
-18 -2x - 7 < 0
Graphs of Inequalities in two variables
1.
x + 3y 9
6. 1/2x + 1/3y + 6 0
2.
8x + 3y 24
7.
2y - x + 1 2x + y - 1
3.
1/2y + x - 4
8.
x + 2 < 1/2 (y + x + 1)
4.
x+y2
9.
x/2 - 1 > y/4 + x/5
5.
x + 3y 9
68
Chapter 5
5.1
QUADRATIC EQUATIONS
Imaginaries
Recall that in our lessons on radicals we do not treat problem on negative
radicands like 4 . An introduction to this concept is hereby introduce.
Imaginaries is a radical with an indicated even roots and whose radicand is
negative as 1, 2 or 3 5 . Imaginary number 1 is known in mathematics
as "i".
Simplifying imaginaries:
4 4 1 4 (1) 2 1 21
7 7 1 i 7
5 6 5 6(1) 5i 6
An imaginary number is considered simplified when one factor is i and the rest
are real numbers.
Exercises:
1
2
2
3
1.
20
6.
2.
12
7.
3. 16
1
4
4
5
8.
3
45
5
1
4
9.
3
4
10. 27
4.
5.
1
5
4
9
69
5.2
Quadratic Equations
Is one in which the unknown letter contains the second power and no higher power.
The general form of a quadratic equation is Ax2 + Bx + C = 0. It may either be complete
containing the first and second powers of the unknown or incomplete when some
powers are lacking as 3x2 - 27 = 0. Every quadratic equation in one unknown has only
two roots.
5.2.1 Forms of Quadratic Equations
1.
Ax2 + Bx + C = 0
2.
Ax2 + c = 0
3.
x2 = C
5.2.2 Solutions to Quadratic Equations:
5.2.2.1 By Factoring
Steps:
1.
2.
3.
4.
Express the given equation in shorthand form.
Factor the left number of the equation.
Equate each factor to zero and solve for the resulting equation.
Check by substituting the solve of the variable in the given equation.
Examples:
1.
4x2 - 100 = 0
(2x - 10)(2x + 10) = 0
2x - 10 = 0, 2x + 10 = 0
x=5
x = -5
2.
x2 - x = -6
x2 - x + 6 = 0
(x - 3)(x + 2) = 6
x = 3, x = -2
Check:
4(5)2 - 100 = 0
4(25) - 100 = 0
100 - 100 = 0
0=0
4(-5)2 - 100 = 0
4(25) - 100 = 0
0=0
70
Exercises:
A.
Quadratic Equations
1.
14x - 11x2 = 3
11.
14
1
1
x2 x4
2.
x2 = 5x
12.
1 x2
x
24
3.
3x2 - 18x = 0
13.
1
1
1
2x 1 2x 1 4
4.
4y2 = 4y - 1
14.
x6
5.
a2 + 5a + 6 = 0
15.
40 20 8
x
x 7
6.
2x 5 1 x 3
7.
x 3 x 5 4
8.
y5 y 5
9.
x 9 x 1
10.
2m 3 m 7 2
Equations Reducible to quadratic
16. x4 - 9x2 + 8 = 0
17. x 3 x 4 0
18. (x2 - 1)2 - (x2 - 1) = 0
19. y-2 - y-1 - 2 = 0
20. x - 10 x 9 = 0
1
6
x
71
B.
Determine the nature of the roots of the equation.
1.
x2 - 6x + 9 = 0
5.
6y2 - 2 3y - 1 = 0
2.
4x2 - 12x + 9 = 0
6.
4x2 - 4 3x + 3 = 0
3.
x2 + 7 = 0
7.
y2 - 1/2y + 3/5
4.
y2 + 9/4 = 4y
5.2.2.2 by Completing the Square
Solving Quadratic Equation by Completing the Square:
This method will make use the concept of perfect trinomial square and finding the
missing term of the perfect trinomial square. The following steps will illustrate the
process.
1.
The equation should be written in the forms ax2 + bx = c and divide both members
by the coefficients of x2 if a is other than 1.
2.
Take ? the coefficient of x and square it. Add this number to both members of the
equation.
3.
Take the square root of both members and write " sign before the right member.
4.
Solve the two linear equation
5.
Check the result.
Example 1. y2 - 4y - 5 = 0
1.
y2 - 4y = 5
4.
y-2=3
2.
y2 - 4y + 4 = 9
5.
y=23
y = 5, y = -1
3.
(y - 2)2 = 9
72
Example 2. 4a2 - 12a - 15 = 0
4a2 - 12a = 15
y2 - 3y = 15/4
y2 - 3y +
9 15 9
4 4 4
(y - 3/2) = 6
y - 3/2 = 6
y = 3/2 6
Exercises: Find the roots by completing the square:
1.
a2 - 12a + 11 = 0
6.
p2 - 4p = 5
2.
2a2 - 5a - 12 = 0
7.
6a2 - 5a = -1
3.
r2 + 18r + 17 = 0
8.
4y2 - 5y = -1
4.
b2 = 21 - 4b
9.
2x2 + 5x = 12
5.
10r2 = -19r - 6
10. 3a2 + 7a = 5
5.2.2.3 by the Quadratic Formula
Solving Quadratic Equation by the Quadratic Formula.
x
b b 2 4ac
2a
The example will illustrate the process.
Example 1. x2 + 5x + 4 = 0 in which a = 1, b = 5 and c = 4 using the above formula
x
b b 2 4ac
2a
=
5 25 4.1.4
2.1
=
5 25 16
2
73
=
5 9
2
=
53
2
x = -1, x = -4
Example 2. x2 + 2x = -5 in which a = 1, b = 2, c = 5
x
2 4 4.1.5
2.1
x=
2 4 20
2
x
2 16
2
x=
2 4i
2a
xi
2 4i
2 4i
, x2
2
2
x i 1 2i, x 2 1 2i
Exercises:
1.
3y2 - 5y = 2
5.
5r2 = 9r + 2
9.
2a2 + a - 15 = 0
2.
15m2 + 14m - 8 = 0
6.
3y2 - y = 2
10. y2 - 2y = 24
3.
4y2 - y - 5 = 0
7.
18y2 - 2 = -24y
4.
6x2 + 7x - 20 = 0
8.
4y2 + 16y + 15 = 0
74
5.2.3 Verbal Problems Using Quadratic Equations
1.
A man wants to enclose his rectangular lot, one side of which is already fenced.
The total length of the fenced is 104 meters. What are the dimension of the lot if
the area is 200 mm?
2.
The square of a number y is 90 more than the number. Find the number.
3.
If a certain number is added to its square he result is 30. Find the number.
4.
The area of a rectangle is 24 cm2. Find the base and altitude if the altitude is 2 cm.
less than the base.
5.
Twice the square of a number increased by nine time the number is 5. Find the
number.
6.
Find the dimension of a rectangle if its length is 6 cm greater than it width and the
area is 40 cm?
7.
Find the area of a piece of square board of its diagonal is 8 m more than the side.
8.
The current of a river is 1 km/hr. If a man takes 8 hrs more to row 12 km upstream
than to row downstream at the same distance. Find his rate in still waters.
9.
The sum of the square of two consecutive number is 145. What are the numbers?
10. Felix has a rectangular garden which is to be fenced along 3 sides. The fourth side
is a high wall of a neighbor. Find the dimension of the lot if it uses 120 meters of
wire for fencing and if its area is 1000 m2?
5.3
Graphs of Quadratic Functions
More generally, the graph of the solution set of any quadratic equation of th form
y = ax2 + bx + c
equations, it is desirable to select first component of the ordered pairs that will insure
that the significant parts of the parabola are displayed. These parts include the
intercept and the highest and lowest points of the curve (vertex).
In practice, it is easier to determine the y-coordinate of the minimum an maximum
point by finding the x-value, x = -b/20 and then substituting this value in the functions.
Since a value of "a" greater than 0 indicates that the curve has a minimum point, it
follows that the parabola opens upward for a > 0. In a similar way, a parabola opens
downward when a < 0.
75
For the function f(x) = ax2 + bx + c, any value of x for which f(x) = 0 will be a
solution to the equation. These are called the x-intercepts on the zeros of the functions.
In graphical solutions, the best that we can expect is an approximation to the actual
solutions. If the equations has integral solutions, we may happen to get exact results,
generally however, we have to estimate the intercepts.
A quadratic equation may have no real solutions, one real solutions or two real
solutions. If the equation has no real solution, the graph of the equation does not touch
the x-axis. It there is one solution, the graph in tangent to the x-axis, if there are two
real solutions, the graph crosses the x-axis in two distinct points.
5.3.1 Sketching the Quadratic Functions:
1.
2.
3.
4.
5.
Find and plot the y-intercept.
Find and plot the coordinate of the minimum or maximum point
Draw the axis of symmetry
Find and plot a point symmetrical to the y-intercept with respect to the axis of
symmetry.
Additional points maybe plotted using the real zeros of the function.
Example: f(x) = x2 + 4x + 1
1.
The y-intercept is (0, 1)
2.
Since a > 0 have maximum the parabola open upward. Vertex is (-2, -3)
3.
The axis of symmetry is -2
Solution: Let x = 0
y=0
x=
b
2a
y = x2 + 4x + 1
= -22 + 4(-2) + 1
=+4-8+1
4
= -2
2(1)
Vertex = -3
Use the quadratic formula to find the zeros of the function.
76
Exercises: Show the graph of the following quadratic equations.
1.
g(x) = -x2 + 3x - 2
5.
f(x) = x2 - 6x + 5
2.
h(x) = -3x2 - 6x - 1
6.
g(x) = 2x2 + 5x - 7
3.
h(x) =
1 2
x - 2x
2
7.
y = -x2 - 8x - 16
4.
f(x) =
1 2
x +2
2
8.
n(x) = -3x2 - 11x + 4
77
Chapter 6
SYSTEMS OF EQUATIONS
Often in our work with problems, the use of one unknown is not enough to yield in
equation as well as the answer to the problem presented. In order to be able to solve
problems in which the use of more than one unknown is desirable, we shall examine
some methods of solving equations in two or more unknowns.
If, we have two linear equation in the same two unknowns, there can usually be
found one and only one set of values, which will satisfy both equations. When two or
more equation are considered together and a common solution for hem is desired, this
is called a system of equation.
Simultaneous systems are sets of equations in the same number of unknown.
Each equation is consistent with the others and independent for them. Such systems
can be solved.
Example: x - y = 5 the roots of the systems are x = 4, y = -1.
A good way to visualize the solution set of simultaneous system is to consider the
graph of the equations. If there are ordered pair that satisfy both equations, then the
points corresponding to these ordered pairs will have to lie on the graphs of both. Since
the graphs of linear equations are straight lines, three possibilities can occur. If the
lines are parallel, there is no common solution because the lines have no points in
common. These are called inconsistent systems. If the lines are identical having the
same graph. They have the same solution set. Such equations are called dependent
systems. If the graphs intersects at one point the equations in such system are
independent and consistent.
6.1
Graphical method
Example: Solve the roots by graphic method
x+y=6
x-y=2
-------------
x-intercept
(6, 0)
(2, 0)
y-intercept
(0, 6)
(0, -2)
The point of intersection is (4, 2) hence the solution set.
6.2
Algebraic Solutions of Simultaneous Equations in Two Unknowns
6.2.1 Elimination by Addition or Subtraction
Steps:
1.
If necessary multiply each side of one or both equations by same numbers so that
the numerical coefficients of one of the variables are of equal absolute value.
78
2.
3.
4.
If the coefficients of equal absolute value have the signs, subtract one equation
from the others. If they have unlike signs added the equations.
Solve the resulting equation for the remaining variable.
Check by substituting the roots in both original equation.
Example: 4x + 5y = 6 (1)
2x + 3y = 1 (2)
-------------12x + 15y = 18 Eq. 1 multiply by 3 - (3)
10x + 15y = 20 Eq. 2 multiply by 5 - (4)
---------------2x
= -2 Subtracting Eq. 4 from 3
x = -1
y=2
by substituting the value of x in Eq. 1
Check: -4 + 10 = 6
-2 + 6 = 4
6.2.2 Elimination by Substitution
Steps:
1.
2.
3.
4.
5.
In any of the two given equations, solve for one variable in terms of the other.
Substitute this resultant value of the variable from step 1 in the remaining equation.
Solve the equation obtained in step 2 for the remaining variable.
From the equation in step 2, substitute the solution for the variable found in step 3
and solve for the resulting equation.
Check by substituting the ordered pair in both original equation.
Example 2: 5x + 3y = 0
8x - y = 29
--------------y = 8x - 29
(1)
(2)
(3) Using Eq. 2 solve for y in term of x
Substitute the value of y in Eq. 1
5x + 3 (8x - 29) = 0
29x = 87
x =3
By substituting x = 3 in Eq. 3
y = -5
79
Check by substituting the ordered pair (3, -5) in both equations
15 - 15 = 0
24 + 5 = 29
Systems of Linear Equations in Three Unknowns
Systems of this type maybe solved by first eliminating one of he unknowns and
thereby reducing the problem to that of solving systems in two unknowns. We proceed
to solve the reduced system of equation as discussed earlier when we have to solve
systems in three unknowns we can eliminate best by either addition or subtraction. An
example will best illustrate the process.
Solve the system
x+y+z=2
2x - 2y - z = 2
x + 2y - z = -3
------------------3x - y = 4
2x + 3y = -1
9x - 3y = 12
2x + 3 = -1
------------------11x = + 11
x=1
(1)
(2)
(3)
(4) add eq. 1 & 2 to eliminate z
(5) add eq. 1 & 3 to eliminate z
using Eq. 4 & 5 eliminate
(6) Multiply by 3
(7) repeat Eq. 5
Solve for y using Eq. 5 by substituting x = 1 hence y = -1
Using the values (1, -1) solve for z from using the original equation, hence z = 2.
To check substitute the values of x, y, and z.
6.2.3 Determinants
Solving Systems of Equations by Determinants
Determinants are square array of numbers written between vertical lines as
a 1 b1
.
a 2 b2
The four quantities involve are called elements of the determinant. A second order
determinants consist of two rows and two columns (2 x 2). The value of the determinant
is obtained by getting the product of the principal diagonal minus the product of the
secondary diagonal. In the case of the above determinant, the product of the principal
diagonal is a1b2 and secondary diagonal is a2b1.
80
a1b2 - a2b1. A third order determinant consist of 3 rows and 3 columns. The
scheme of diagonal multiplication will aid in remembering the expressions. In case of
third order determinants the first two columns are repeated at the right.
Example 1. Solve the linear system by determinant: 2x + y = 4, x - 3y = -5
Solution:
4
x
1
5 3
2
1
y
3
1
2
4
1
2
5
1
1
3
Notice that the determinants form the denominators of x and y. It is formed by the
coefficients of x and y terms in the order in which they stand. The determinants in the
numerator in each case is the same as that in the denominator except that he numerical
terms replace the coefficients of the letters for which we are solving. The example
above can be simplified as:
x=
12 (5)
7
1
6 1
7
y=
10 4 14
2
6 1
7
therefore x = 1, y = 2 is the solution set.
Example 2: Solve the given system by determinants
x+y+z=1
x + 3y + 2z = 8
2x + 8y - 3t = 15
x
1
1
1 1
1
8
3
2 8
3
15
1
8
1
3 15
1 1
8
1
1
3
2 1
3
2
8
3 2
8
x
9 30 64 45 16 24 48
3
9 4 8 6 16 3
16
81
y
z
1
1
1 1
1
1
8
2 1
8
2
1
15
1
3 2
1 1
15
1
1
3
2
1
3
2
8
3 2
8
1
1
1 1
1
1
3
8 1
3
2
1
8
1
15 2
1 1
8
1
1
3
2 1
3
2
8
3 2
8
y=
24 4 15 16 30 8 48
3
16
16
z
45 16 8 6 64 15 16
1
16
16
Hence x = -3, y = 3, and z = 1. In all cases of division the denominators determinants
must not be 0.
The methods of elimination, substitution and determinants can also be use to
solved non-linear systems.
Exercises: Find the solution sets of the systems.
1.
-4x + 3y = 0
8x - 6y = 0
------------------
2.
x-y =6
x2 + y2 = 26
------------------
4.
y = x2
y=x+2
--------------
5.
y = x2
y=x-?
-------------
7.
x2 - y2 = 7
2x2 + 3y2 = 24
-------------------
8.
4x2 + y2 = 25
x2 + 4y2 = 40
-------------------
10.
x - 5y + 2z = 0
-2x - y + z
=0
x - 16y + 7z = 0
-------------------------
11.
3. x - y = 1
xy = 12
----------------6.
x - 2y + z = 4
3y + 2z = -2
4y - z = 1
----------------------
x2 + y2 + y = 10
x+y+1 =0
---------------------9. x2 + y2
=5
x2 - 2xy + y2 = 1
------------------------12. 6x - 7y + 5z = 0
3x + 4y - 2z = 0
x-y-z
=0
-----------------------
82
Problem Application:
1.
The width of a rectangle is 2/3 of its length and its perimeter is 80 meters. Find the
dimension of the rectangle.
2.
The sum of the digit of a certain integer having two digits is 9. If the digits is
reversed, the new number is 9 less then 3 times the original number. Find the
numbers.
3.
The sum of three numbers is 12. Twice the first number is equal to the second and
the third number is equal to the sum of the other two. Find the numbers.
4.
A man rowed 10 km up the river in 5 hours and back in 2 hrs. Find the rate of the
current and his rate of rowing in still waters.
5.
The sum of two integers is 84 and their difference is 60. Find the integers.
6.
Ofelia has P8.50 in 2.5 cts and 50 cts piece. If she has 25 coins in all, how many of
each kind has she.
7.
8 times a certain number added to 5 times a record number is 184. The first
number is three more than the second number.
8.
Alice is twice as old as her sister Merle. Four years ago, she was three times as
old as Merle. How old is each girl now?
9.
Deling receive P20.00 to buy 6 cans of milk and two bars of soap expecting a
change of 2.80. She made a mistake, instead she bought 4 cans of milk and 3 bars
of soap with a change of P2.20. How much was the cost of a can of milk and a bar
of soap.
A.
Graphing:
1.
x-y<2
x+y>4
5.
3x + y > 6
x+y<0
2.
3x - 2y 3
2x - 3y 5
6.
6x - 2y < 12
y-3<0
x+y>0
3.
x+y>1
x + y < -3
7.
8x + 5y 40
x + 26 8
x0
y0
4.
x + y > -2
y-z<2
8.
y 2x + 1
y -2x + 1
x2
83
TABLE OF CONTENTS
CHAPTER 1
ALGEBRAIC EXPRESSIONS .................................................................................................1
1.1
INTRODUCTION ............................................................................................................................................1
1.2
GROUPING ALGEBRAIC EXPRESSIONS .........................................................................................................2
1.3
LAWS OF EXPONENTS ..................................................................................................................................4
1.4
OPERATIONS ON ALGEBRAIC EXPRESSIONS ................................................................................................8
1.4.1
Addition/Subtraction of Algebraic Expressions ..............................................................................8
1.4.2
Multiplication of Algebraic Expressions ....................................................................................... 10
1.4.3
Division of Algebraic Expressions ................................................................................................. 11
1.5
SPECIAL PRODUCTS AND FACTORING ........................................................................................................ 16
1.5.1
The Product of a Monomial and a Polynomial ............................................................................. 16
1.5.2
The Product of Binomials Having Dissimilar Terms ................................................................... 17
1.5.3
The Product of Two Binomials Having Similar Corresponding Terms ....................................... 18
1.5.4
The Square of a Binomial .............................................................................................................. 20
1.5.5
The Product of the Sum and Difference of the Same Binomials ................................................. 22
1.5.6
The Cube of a Binomial ................................................................................................................. 23
1.5.7
The Square of a Polynomial .......................................................................................................... 24
1.5.8
Others Types of Factoring ............................................................................................................. 25
1.6
RATIONAL EXPRESSIONS ........................................................................................................................... 28
1.6.1
Simplification of Fractions ............................................................................................................ 30
1.6.2
Addition and Subtraction of Fractions .......................................................................................... 31
1.6.3
Multiplication of Fractions ............................................................................................................ 34
1.6.4
Division of Fractions...................................................................................................................... 35
1.6.5
Complex Fractions ......................................................................................................................... 36
1.7
RADICAL EXPRESSIONS ............................................................................................................................. 38
1.7.1
Simplification of Radical Expressions........................................................................................... 40
1.7.2
Rationalizing Radical Expressions ................................................................................................ 40
1.7.3
Multiplication and Division of Radical Expressions. ................................................................... 43
1.7.4
Addition and Subtraction of Radical Expressions ........................................................................ 44
CHAPTER 2
RELATION AND FUNCTIONS............................................................................................. 48
2.1
RELATION.................................................................................................................................................. 48
2.1.1
Picturing Relations: ....................................................................................................................... 48
2.2
FUNCTION ................................................................................................................................................. 50
CHAPTER 3
LINEAR EQUATIONS ........................................................................................................... 55
3.1
SOLUTION OF EQUATIONS ......................................................................................................................... 55
3.2
LINEAR EQUATION IN ONE UNKNOWN: ......................................................................................... 56
3.3
LINEAR EQUATION IN TWO UNKNOWNS ....................................................................................... 58
3.3.1
GRAPH OF LINEAR EQUATION IN TWO UNKNOWNS ........................................................ 59
3.3.2
SLOPES OF THE LINE ............................................................................................................... 59
3.3.3
FORMS OF LINEAR EQUATION .............................................................................................. 59
CHAPTER 4
4.1
4.2
4.3
4.4
INEQUATIONS ....................................................................................................................... 63
ADDITION PRINCIPLE OF AN INEQUALITY .................................................................................................. 63
MULTIPLICATION PRINCIPLE FOR INEQUALITIES ....................................................................................... 63
GRAPHS OF INEQUATION ........................................................................................................................... 65
COMPOUND INEQUALITIES ........................................................................................................................ 66
CHAPTER 5
QUADRATIC EQUATIONS .................................................................................................. 68
5.1
IMAGINARIES ............................................................................................................................................. 68
5.2
QUADRATIC EQUATIONS ........................................................................................................................... 69
5.2.1
Forms of Quadratic Equations ...................................................................................................... 69
5.2.2
Solutions to Quadratic Equations: ................................................................................................ 69
5.2.3
Verbal Problems Using Quadratic Equations ............................................................................... 74
5.3
GRAPHS OF QUADRATIC FUNCTIONS ......................................................................................................... 74
5.3.1
Sketching the Quadratic Functions: ............................................................................................. 75
84
CHAPTER 6
SYSTEMS OF EQUATIONS .................................................................................................. 77
6.1
GRAPHICAL METHOD ................................................................................................................................. 77
6.2
ALGEBRAIC SOLUTIONS OF SIMULTANEOUS EQUATIONS IN TWO UNKNOWNS.......................................... 77
6.2.1
Elimination by Addition or Subtraction ........................................................................................ 77
6.2.2
Elimination by Substitution ........................................................................................................... 78
6.2.3
Determinants .................................................................................................................................. 79
