Download Worksheet 17 (4

Worksheet 17 (4.1)
Chapter 4 Polynomials
4.1 Polynomials: Sums and Differences
Summary 1:
Monomials are terms that contain variables with only nonnegative integers
A polynomial is a monomial or a finite sum or difference of monomials.
A binomial is a polynomial with two terms. EX. 5x - 7
A trinomial is a polynomial with three terms. EX. 2x2 - 3x + 6
The degree of a monomial is the sum of the exponents of the literal factors.
EX.
2 4
-12a b c is of degree 7
Any nonzero constant term is of degree zero. EX. -8 is of degree zero
The degree of a polynomial is the degree of the term with the highest
degree in the polynomial. EX. 4x3y4 - 5xy4 is of degree 7
Warm-up 1. a) The degree of 8x2y3 is
.
3
2
b) The degree of 2x - 5x + 3x - 6 is
Problems
1. Determine the degree of 5x2y3 - 3xy3 - 6x3.
2. Determine the degree of 5a - 6b.
Summary 2:
72
.
as exponen
ToTosubtract
add polynomials,
polynomials,
rearrange,
change regroup
the problem
and combine
to addition
similar
of theterms.
opposite.
(To take the opposite of a polynomial, change the signs of all the terms.
Warm-up 2. a) Add 2x2 - 5x + 6 and 3x2 + x - 4.
(2x2 - 5x + 6) + (3x2 + x - 4) = (2x2 + ) + (-5x +
=
x2 x+
) + (6 +
)
b) Find the sum: (2a - 5) + (a + 6) + (3a - 2).
(2a - 5) + (a + 6) + (3a - 2) = (2a +
=
+
a-
) + (-5 +
+
)
Worksheet 17 (4.1)
Problems
3. Add 3y2 - 6y + 2 and y2 - 2y - 5.
4. Find the sum: (5b - 6) + (b - 4) + (2b + 3)
Warm-up 3. a) Subtract 8x2 + 2x - 6 from -3x2 + x - 4.
(-3x2 + x - 4) - (8x2 + 2x - 6) = -3x2 + x - 4 2
= (-3x ) + (x =
x2 x+
+
) + (-4 +
b) Simplify: 6x - [2x - (5x + 2)] = 6x - [2x ]
= 6x - [
x]
= 6x +
x+
=
x+
Problems
5. Subtract -3x2 - 5x + 2 from 2x2 + 3x - 4.
73
)
6. Simplify: -3x + [8x - (7x + 9)].
Polynomials occur in various geometric problems: perimeter, area, volume.
Summary 3:
Warm-up 4. a) Write a polynomial to represent the perimeter of a rectangle whose
width is 2x + 3 and whose length is 5x - 4. Simplify the
expression by combining similar terms.
P = 2W + 2L
P = 2(
) + 2(
P=
x+
+
P=
x-
)
x-
Worksheet 17 (4.1)
b) Find the perimeter of the rectangle in part a if the value of x is 4.
P=
P=
P=
(4) -
The perimeter of the rectangle is
.
Problems
7. Write a polynomial to represent the perimeter of a square with sides of length
2x2 - 5.
8. Find the perimeter of the square in problem # 7 if x = 3.
74
Worksheet 18 (4.2)
4.2 Products and Quotients of Monomials
Summary 1:
Properties of Exponents
If a and b are real numbers and m and n are positive integers, then:
bn  b m = bn + m
(bn)m = bmn
(ab)n = an bn
n
b
= bn- m ; when n > m; (b  0)
m
b
n
b
= 1 ; when n = m; (b  0)
m
b
75
Warm-up 1. Simplify:
a) y8  y4 =
e) (x5)2 =
b) x  x5 =
f) (4x)2 =
9
x
g) 5 =
x
3
a
h) 3 =
a
c) 52 53 =
d) (23)4 =
Problems - Simplify:
1. a5  a9
3. (y8)3
7
m
5. 2
m
Summary 2:
2. 7  78
4. (2y)3
3
5
6. 3
5
To multiply monomials, multiply their coefficients and multiply the variable
factors by adding the exponents of the like variables.
Worksheet 18 (4.2)
Warm-up 2. Find each of the following products:
a) (2x2y2)(4xy4) = (24)(x2x)(y2y4)
=(
)(x2 + 1)(y2 + 4)
76
=
b) (-5x)(3x2)(2x4) = (-532)(xx2x4)
= ( )( )
=
c) (-3x2y)3 = (
=
)3(
)3(
)3
Problems - Find each of the following products:
7. (cd2)(-3c4d5)
8. (2d)(-3d3)(½d5)
9. (2x5y2)4
Summary 3:
To divide two monomials, divide the coefficients and divide the variable
factors by subtracting the exponents of the like variables.
Warm-up 3. Find each of the following quotients:
- 42 a 2 b5  42   a 2   b5 
a)
=  -     3 
6 ab3
 6  a b 
= ( )(
)(
)
=
18 x y  18   x7   y4 
b)
=   
2 x3 y4  2   x3   y4 
= ( )(
)(
)
7
4
=
Problems - Find each of the following quotients:
100 x5 y
- 12 a7 b6
10.
11.
50 x 2 y
3 a5 b
Worksheet 19 (4.3)
4.3 Multiplying Polynomials
77
Summary 1:
To multiply a monomial and a polynomial, multiply the monomial times
each term of the polynomial by using an extension of the distributive property.
Warm-up 1. a) -5x2y(2x2y - 3y2 + 5) = (-5x2y)(2x2y) - (-5x2y)(3y2) + (-5x2y)(5)
=
+
-
Problems - Multiply:
1. 3y2(2y2 - 5y + 9)
2. -xy3(3x2y - 7xy2 + 2)
Summary 2:
To multiply any two polynomials, multiply each term of the first polynomial
times each term of the second polynomial. Simplify by combining
similar
terms if needed.
Warm-up 2. a) (x - 5)(x + 6) = x(x + 6) - 5(x + 6)
= x2 +
- 30
= x2 +
- 30
b) (x + 3)(x2 - 2x + 5) = x(x2 - 2x + 5) + 3(x2 - 2x + 5)
= x3 - 2x2 + 5x +
+
= x3 +
+
Problems - Multiply:
3. (c - 3d)(c + 5a)
78
4. (a - 2)(a2 + 3a - 9)
Worksheet 19 (4.3)
Summary 3:
When multiplying two binomials, a shortened version can be used:
1. Multiply the first terms of the two binomials. (First)
2. Multiply the two outer terms and multiply the two inner terms and
combine if products are similar terms. (Outer + Inner)
3. Multiply the last terms of the two binomials. (Last)
This method is sometimes called the FOIL method.
Warm-up 3. a) (x - 3)(x + 4) =
+
Step 1
Step 2
Step 3
Step 1. Multiply xx
Step 2. Multiply x4 and -3x and combine
Step 3. Multiply -34
b) (2x - 5)(3x + 4) =
-
Problems - Multiply:
5. (x + 6)(x - 7)
6. (5x - 6)(2x + 4)
Summary 4:
79
-
To square a binomial you can multiply the
(a -binomial
b)2 = a2times
- 2abitself
+ b2or you can
2
2
The pattern
follows
the format
use the
following
specialof:
patterns: (a + b) = a + 2ab + b2
(first term)2 ± 2(first term)(second term) + (second term)2
If the sign between the two terms is positive, the middle sign of the answer
will be positive. If the sign between the two terms is negative, the middle
sign of the answer will be negative.
When two binomials differ only in the middle sign a special pattern develops: (a
+ b)(a - b) = a2 - b2
(The sum of the outer and inner products is 0: ab - ab = 0)
When cubing a binomial the following patterns develop:
(a + b)3 = a3 + 3a2b + 3ab2 + b3
(a - b)3 = a3 - 3a2b + 3ab2 - b3
sheet 19 (4.3)
Warm-up 4. Multiply using the special patterns:
a) (x - 5)2
= (x)2 - 2(x)(5) + (5)2
=
+
b) (3a + 2b)2 = (3a)2 + 2(3a)(2b) + (2b)2
=
+
+
)2 - (
-
c) (2x - 7)(2x + 7) = (
=
d) (x - 5)3 = (
=
=
e) (2x + 3)3 = (
=
=
)2
)3 - 3( )2( ) + 3( )( )2 - (
- 3( )( ) + 3( )( ) +
)3 + 3( )2( ) + 3( )( )2 + (
+ 3( )( ) + 3( )( ) +
+
+
+
Problems - Use special patterns to multiply:
7. (5x - 6)2
80
)3
)3
8. (7x - 2)(7x + 2)
9. (y + 3)3
10. (3a - 4b)3
 Note:
If the patterns are forgotten or are difficult for you to use, using the distributive property
will always work.
Worksheet 20 (4.4)
4.4 Factoring: Use of the Distributive Property
Summary 1:
A prime number is a positive integer greater than 1 that has no factors that
are positive integers other than itself and 1. EX. 2, 3, 5, 7
A composite number is a positive integer greater than 1 that is not a prime
number. EX. 4, 6, 8, 9
Every composite number is the product of prime numbers. EX. 6 = 23
Prime factorization form of a number is the indicated product form that
contains only prime factors. EX. 63 = 337
A number is completely factored when it is in the prime factorization form.
81
Warm-up 1. Classify as prime or composite:
a) 92
92 = 192, 246 or 423 therefore is
b) 47
47 = 147 therefore is
.
.
Problems - Classify as prime or composite:
1. 33
2. 79
3. 129
Warm-up 2. Factor into the product of prime numbers:
a) 42 = 2
= 23
b) 140 = 2
= 22
= 225
Problems - Factor into the product of prime numbers:
4. 92
5. 125
6. 68
We use the distributive property, ab + ac = a(b + c), to factor a polynomial
whose terms have common factors.
EX. 5x - 10y = 5(x) - 5(2y) = 5(x - 2y)
This type of factoring process is referred to as factoring out the highest common
monomial factor.
A polynomial with integral coefficients is in completely factored form if:
1. It is expressed as a product of polynomials with integral coefficients.
2. No polynomial, other than a monomial, within the factored form can
be further factored into polynomials with integral coefficients.
Summary 2:
Worksheet 20 (4.4)
Warm-up 3. Factor completely:
a) 15x3 - 5x2 = 5x2(
82
)
b) 3x2y3 + 12xy2 = 3xy2(
+
)
c) 5(a - b) - x(a - b) = (a - b)(
)
 Note: Warm-up c) is an example of a common binomial factor.
Problems - Factor completely:
7. 24a4b - 8a3b2
8. 18x3 + 12x2 - 24x
9. y(a - 2) + 5(a - 2)
10. x2(y + 5) - 8(y + 5)
Summary 3:
Factoring by grouping is usually used when there are four terms.
To factor by grouping:
1. group the four terms in pairs.
2. factor out the common factor for each pair.
3. factor out the common binomial factor which appears in step 2.
Warm-up 4. Factor by grouping:
a) ax + bx + ac + bc = x( +
= (x + c)(
b) 2y - 4z + ay - 2az = 2( =( +
c) y2 - 3y - 4y + 12 = y(
=(
-
) + c(
+ )
) + a(
)( -
) - 4(
)( -
+
)
)
)
)
)
 Note: It was necessary to factor -4 from the last two terms for the binomial factors to match.
d) ac - bd + bc - ad
= ac - ad + bc - bd
83
= a(
=(
+
) + b(
)( -
)
)
 Note: It was necessary for this polynomial to be rearranged since no common factor occurred in
the two pairs as it was written.
Worksheet 20 (4.4)
Problems - Factor by grouping:
11. 2x - ax + 6 - 3a
12. ax + bx - ay - by
13. 5m - cn + 5n - cm
Summary 4:
The following property allows us to use factoring as a technique for solving
equations:
Let a and b be real numbers, ab = 0 if and only if a = 0 or b = 0
This property tells us that if two factors multiply to give 0, one of the factors
must be equal to 0.
y2 + 8y = 0
y( + ) = 0
+
=0
y = 0 or y =
Solution Set = { 0,
4c2
4c2 - 12c
4c( - )
4c = 0 or
c =
or c
= 12c
=0
=0
=0
=
Solution Set = {
Warm-up 5. a) Solve:
y = 0 or
b) Solve:
c) Solve 5ax2 - 10cx = 0 for x:
84
}
,
}
5ax2 - 10cx = 0
5x( - ) = 0
5x = 0 or
=0
x=
or
x =
x=
Solution Set = {
,
}
Worksheet 20 (4.4)
Problems - Solve:
14. 2x2 - 4x = 0
15. 5m2 = 15m
16. Solve for y: 2ay2 - by = 0
Warm-up 6. Set up an equation and solve:
a) The area of a square is 5 times its perimeter. Find the length of a
side of the square.
A = s2 P = 4s (Geometry formulas needed.)
A
s2
s2
s2 s(s s = 0 or s s=0
or
= 5(P)
= 5(4s)
= 20s
=0
) =0
=0
s =
The length of a side of the square is
.
Problems - Set up an equation and solve:
17. The area of a circular region is numerically equal to four times the circumference
of the circle. Find the length of the radius of the circle.
85
Worksheet 21 (4.5)
4.5 Factoring: Difference of Two Squares and
Sum or Difference of Two Cubes
Summary 1:
Difference of Two Squares: a2 - b2 = (a + b)(a - b)
A sum of two squares will not factor. It is said to be a prime polynomial.
a2 + b2 is a prime polynomial
Warm-up 1. a) x2 - 36 = (
=(
)2 - (
)2
)(
b) 25a2 - 49 = (
=(
)2 - (
c) a4 - 16b4 = (
=(
=(
=(
)2 - (
+
+
)
)2
)(
)
)2
)(
)[(
)(
)
2
) - ( )2]
)(
d) (a - 1)2 - b2 = [(a - 1) + ][(a - 1) =(
)(
)
e) 2x2 - 18 = 2(
= 2[(
= 2(
)
2
) - ( )2]
)(
f) x2 + 25 =
86
)
)
)]
Problems - Factor completely:
1. y2 - 64
2. 16a2 - 9
3. x4 - 81
4. (x + 2)2 - y2
5. 3x2 - 48
6. a2 + 81
Worksheet 21 (4.5)
Summary 2:
Sum of Two Cubes: a3 + b3 = (a + b)(a2 - ab + b2)
Difference of Two Cubes: a3 - b3 = (a - b)(a2 + ab + b2)
Warm-up 2. a) x3 - 8 = (
=(
=(
)3 - (
-
b) 27x3 + 1 = (
=(
=(
)3
)[(
)(
)2 + (
)3 + ( )3
+
)[(
)(
c) 125y3 - 64z3 = (
=(
=(
)(
)+(
)2]
)
)2 - (
)3 - (
)3
)[(
)(
)(
)2 + (
)+(
)
)(
)2]
)+(
)
Problems - Factor using the sum or difference of two cubes pattern:
7. a3 - 27
87
)2]
8. 8x3 + 125
9. 64x3 - 1
Summary 3:
Special kinds of equations can be solved using the difference of two squares
factoring pattern. (The two terms must be perfect squares, and the x term
must be missing.)
Warm-up 3. a) Solve:
(
25x2 = 49
25x2 =0
+ )( - ) = 0
+
= 0 or
=0
x=
or
x =
x=
or x =
Solution Set = {
,
}
Worksheet 21 (4.5)
b) Solve:
(
+
2x2 - 18 = 0
2(
) =0
(
) =0
+
)(
) =0
= 0 or
=0
x=
or x =
Solution Set = {
,
}
x3 - 36x
x(
)
x(
+
)(
)
x = 0 or
+ = 0 or x=0
or
x=
or x
c) Solve:
Solution Set = {
88
,
,
=0
=0
=0
=0
=
}
d) A rectangle is three times as long as it is wide and its area is 75
square feet. Find the length and width of the rectangle.
Let x = width and 3x = length
widthlength = Area
( )( ) = 75
x2 = 75
x2 =
2
x =0
(x + )(x - ) = 0
= 0 or
=0
x=
or x =
3x = 3( ) =
The width is
and the length is
.
 Note: The solution x = -5 was excluded because the width can not be a negative number.
Problems - Solve:
10. x2 - 81 = 0
Worksheet 21 (4.5)
11. 4x2 = 49
12. x3 - 100x = 0
89
13. The cube of a number equals four times the square of the same number. Find the
number.
Worksheet 22 (4.6)
4.6 Factoring Trinomials
Summary 1:
Trinomials of the Form x2 + bx + c
Factorable trinomials such as x2 + 5x + 6 will factor into the product of two
binomials; x2 + 5x + 6 = (x + 2)(x + 3), where:
1. The first terms of the two binomials multiply to give x2, the first
term of the trinomial. (xx = x2)
2. The second terms of the two binomials multiply to give 6, last
term of the trinomial. (23 = 6)
3. The second terms of the two binomials add to give 5, the
coefficient of the middle term of the trinomial. 2 + 3 = 5
90
Warm-up 1. Factor:
 Note:
If the last sign of the trinomial is positive, the signs in the middle of the two factors will be
alike, and will be the same as the sign in the middle of the trinomial.
a) x2 + 8x + 15 = (x +
)(x +
)
 
both
positive
 Note:
signs
alike
The two numbers in the blanks must multiply and give 15, and add to give 8.
b) x2 - 9x + 20 = (x  
)(x -
)
both signs
negative alike
 Note: The two numbers in the blanks must multiply and give 20, and add to give - 9.
 Note: If the last sign of the trinomial is negative, the signs in the middle of the two factors will be
different. The sign in the middle of the trinomial will go in front of the larger of the two numbers
used for the second terms.
c)
x2 - x - 12 = (x +

)(x -
)
negative signs
in front of different
larger number
 Note:
The two numbers in the blanks must multiply and give -12, and add to give -1.
Worksheet 22 (4.6)
d)
x2 + 3x - 18 = (x +
 
)(x -
)
positive signs
in front of different
larger number
 Note:
The two numbers in the blanks must multiply and give -18 and add to give 3.
91
Problems - Factor completely:
1. x2 - 7x + 10
2. x2 + 5x - 24
3. y2 + 13y + 42
4. a2 - 5a - 14
Summary 2:
Trinomials of the Form ax2 + bx + c
Factorable trinomials such as 2x2 - x - 10 will factor into the product of two
binomials; 2x2 - x - 10 = (2x - 5)(x + 2), where:
1. The first terms of the two binomials multiply to give 2x2, the
first term of the trinomial. (2xx = 2x2)
2. The last terms of the two binomials multiply to give -10, the
last term of the trinomial. (-52 = -10)
3. The products of the inner and outer terms of the two binomials
combine to give the middle term of the trinomial.
(-5x + 2x2 = -5x + 4x = -x)
92
Worksheet 22 (4.6)
Warm-up 2. Factor:
a)
3x2 - 5x - 2 = (3x
 
the inner and outer
products add to
give a negative
)(x
)
signs
different
 Note:
The two numbers in the blanks must multiply and give -2, and the products of the inner
and outer terms must add to give -5x.
b)
6x2 + x - 12 = (2x
)(3x
)
 
the inner and outer signs
products add to different
give a positive
 Note:
The two numbers in the blanks must multiply and give -12, and the products of the inner
and outer terms must add to give +x. If 6x and 1x had been chosen for the first terms, then the
product of the inner and outer terms would not have added to give +x.
c)
10x2 - 19x + 7 = (5x
 
)(2x
)
both signs
negative alike
 Note:
The two numbers in the blanks must multiply and give 7, and the products of the inner
and outer terms must add to give -19x. If 10x and 1x had been chosen for the first terms, then the
product of the inner and outer terms would not have added to give -19x.
Problems - Factor completely:
5. 4x2 - 5x - 6
6. 5x2 - 22x + 8
93
7. 12x2 + 4x - 5
Worksheet 22 (4.6)
Summary 3:
Perfect Square Trinomials
Trinomials in the form of a2 + 2ab + b2 are called perfect square trinomials
since they factor into two identical factors which can be written as the
square of a binomial:
a2 + 2ab + b2 = (a + b)(a + b) = (a + b)2
a2 - 2ab + b2 = (a - b)(a - b) = (a - b)2
Look for this pattern:
(lst term)2 ± 2(lst term)(2nd term) + (2nd term)2 = (lst term ± 2nd term)2
Warm-up 3. Factor using the perfect square trinomial pattern:
a) x2 - 18x + 81 = (x)2 -2(x)(9) + (9)2
=(
)2
b) 16x2 + 24x + 9 = (
=(
)2 + 2( )(
+
)2
)+(
)2
Problems - Factor using the perfect square trinomial pattern:
8. a2 + 8a + 16
 Note:
9. 4x2 - 20x + 25
Trial and error will also work on perfect square trinomials.
94
Summary 4:
Review of Factoring Techniques
1. Factor out a common monomial (or binomial) factor by using the
distributive property. (Always look for this type of factoring first.)
2. If the polynomial contains 4 terms, try factoring by grouping.
3. If the polynomial contains 2 terms, look for: a difference of two squares
pattern, a difference of two cubes pattern or a sum of two cubes
pattern.
4. If the polynomial contains 3 terms, look for a perfect square trinomial
pattern or factor by trial and error.
5. Always check to see if any factor can be factored further.
Worksheet 22 (4.6)
Problems - Factor completely by any method discussed in this chapter.
10. x3 + 27
11. ax - bx + ay - by
95
12. 5x2y3 - 10x3y
13. 9x2 - 12x + 4
14. 81y2 - 25
15. 6b2 + 7b - 20
Worksheet 23 (4.7)
4.7 Equations and Problem Solving
Summary 1:
In this section we continue to use the following property to solve equations:
ab = 0 if and only if a = 0 or b = 0
96
Warm-up 1. a) Solve: 10x2 - x - 3 = 0
10x2 - x - 3 = 0
(5x )(2x +
)
5x = 0 or 2x +
5x =
or 2x
x=
or x
Solution Set= {
=0
=0
=
=
,
}
b) Solve: -2x2 + 6x + 20 = 0
x-
-2x2 + 6x + 20
-2(x2 (x2 )
(x )(x +
)
= 0 or x +
x=
or x
Solution Set = {
=0
=0
=0
=0
=0
=
,
}
c) Solve: (x + 2)(x - 1) = 10
(x + 2)(x - 1)
x2 + x - 2
x2 + x
(x +
)(x )
x+
= 0 or x x=
or x
Solution Set = {

= 10
= 10
=0
=0
=0
=
,
}
Note: In this problem it was necessary to simplify and set = 0 before using the factoring
technique to solve.
Worksheet 23 (4.7)
Problems - Solve:
1. 12x2 + 8x - 15 = 0
97
2. -2a2 - 14a + 60 = 0
3. x(x - 5) = 6
Summary 2:
Many application problems can be solved by using factoring techniques
Warm-up 2. a) Find two consecutive even whole numbers whose product is 120.
Let x = first consecutive even whole number
x + 2 = second consecutive even whole number
x(
+
) = 120
x2 +
= 120
2
x +
=0
(x +
)(x ) =0
x+
= 0 or x =0
x=
or x =
x+2 =
+2=
The two consecutive even whole numbers are
 Note:
.
The negative solution for x was excluded because the problem specified a whole number.
Worksheet 23 (4.7)
Problems - Solve:
4. Find two consecutive odd whole numbers whose product is 63.
98
5. The length of a rectangle is 5 inches more than twice the width. If the area of the
rectangle is 133 square inches, find the width and length.
Summary 3:
Pythagorean Theorem
The Pythagorean Theorem is used in application problems involving right
triangles. The Pythagorean Theorem states: In any right triangle, the
square of the hypotenuse (the longest side, opposite the right angle),
equal to the sum of the squares of the other two sides (called legs).
The Pythagorean Theorem is sometimes stated: a2 + b2 = c2
where a and b represent the legs and c represents the hypotenuse.
99
is
Worksheet 23 (4.7)
Warm-up 3. a) The length of one leg of a right triangle is 2 inches more than the
length of the other leg. If the length of the hypotenuse is 10 inches,
find the lengths of the two legs.
Let:
x = length of shorter leg (a)
x + 2 = length of longer leg (b)
10 = length of the hypotenuse (c)
a2 + b2 = c2
x 2 + (x + 2)2
x2 +
+
+
=
2
2x +
2(
+
)
(
+
)
(x +
)(x )
x+
= 0 or x x=
or x
x+2
The two legs are
and
= 102
=0
=0
=0
=0
=0
=
=
.
Problem - Solve:
6. The length of one leg of a right triangle is 4 cm more than the length of the other
leg. The length of the hypotenuse is 4 cm more than the length of the longer leg.
Find the length of the three sides of the triangle.
100
101