Download Math 101 Study Session Spring 2016 Test 4 Chapter 10, Chapter 11

Math 101 Study Session Spring 2016 Test 4 Chapter 10,
Chapter 11 Chapter 12 Section 1, and Chapter 12
Section 2
April 11, 2016
Chapter 10 Section 1: Addition and Subtraction of Polynomials
A monomial is a number, a variable, or a product of numbers and variables.
A polynomial is a variable expression in which the terms are monomials.
A polynomial of two terms is a binomial.
A polynomial of three terms is a trinomial.
The degree of a polynomial in one variable is the value of the largest exponent on the
variable.
The degree of 4x3 − 3x2 + 6x − 1 is 3, the degree of 5y 4 − 2y 3 + y 2 − 7y + 8 is 4, the degree
of a nonzero constant (number) is 0, and the number zero has no degree.
Polynomials can be added using a vertical or horizontal format.
The opposite of a polynomial is the polynomial with the sign of every term changed, i.e.,
−(x2 − 2x + 3) = −x2 + 2x − 3.
Chapter 10 Section 2: Multiplication of Monomials
Rule for Multiplying Exponential Expressions
If m and n are integers,then xm · xn = xm+n .
Rule for Simplifying Powers of Exponential Expressions
If m and n are integers, then (xm )n = xmn
Rule for Simplifying Powers of Products
If m, n, and p are integers, then (xm y n )p = xmp y np
1
Chapter 10 Section 3: Multiplication of Polynomials
FOIL Method
(A + B) (C + D) = AC + AD + BC + BD
Product of the Sum and Difference of Two Terms
(a + b) (a − b) = a2 − b2
The Square of a Binomial
(a + b)2 = (a + b) (a + b) = a2 + 2ab + b2
(a − b)2 = (a − b) (a − b) = a2 − 2ab + b2
Chapter 10 Section 4 Integer Exponents and Scientific
Notation
Zero as an Exponent
If x 6= 0, then x0 = 1. The expression 00 is not defined.
Definition of Negative Exponents
1
1
If n is a positive integer and x 6= 0, then x−n = n and −n .
x
x
Rules for Dividing Exponential Expressions
xm
If m and n are integers and x 6= 0, then n = xm−n .
x
Chapter 10 Section 5: Division of Polynomials
We divide polynomials using a method similar to the method used for the division of whole
numbers.
2
Figure 1: Dividing Whole Numbers
Figure 2: Dividing Polynomials
Chapter 11 Section 1: Common Factors
The greatest common factor (GCF) of two or more monomials is the greatest integer
and the variable with the smallest exponent that is a factor of each monomial. For example
the GCF of x2 , x4 , and x6 is x2 because 2 is the smallest exponent of the three monomials.
To factor a polynomial means to write the polynomial as a product of other polynomials.
Chapter 11 Section 2: Factoring Polynomials of the
Form x2 + bx + c
To factor polynomials of the form x2 + bx + c we ask ourselves “What two numbers multiply
to give c AND add to give b?” In other words, we seek integers m and n such that m · n = c
and m + n = b. Then the polynomial x2 + bx + c can be factored as (x + m) (x + n). Thus,
x2 + bx + c = (x + m) (x + n).
3
Chapter 11 Section 3: Factoring Polynomials of the
Form ax2 + bx + c
One method we can use to factor polynomials of the form ax2 + bx + c is called the trial and
error method. To use the trial and error method, we use the factors of a and the factors of c.
Figure 3: Trial and Error Method
Another (faster) method we can use to factor polynomials of the form ax2 + bx + c is
called the ac method or the factor by grouping method. To use factor by grouping, we
multiply a · c and look for two numbers m and n so that a · c = m · n and m + n = b. Then
we will split the middle term into mx + nx so that the polynomial ax2 + bx + c becomes
ax2 + mx + nx + c. Remember, we can only use factor by grouping on a polynomial that
has four terms.
Chapter 11 Section 4 Special Factoring
Factoring the Difference of Two Squares
a2 − b2 = (a + b)(a − b)
Factoring a Perfect-Square Trinomial
a2 + 2ab + b2 = (a + b) (a + b) = (a + b)2
a2 − 2ab + b2 = (a − b) (a − b) = (a − b)2
4
Factoring the Sum or Difference of Two Perfect Cubes
a3 + b3 = (a + b) a2 − ab + b2
a3 − b3 = (a − b) a2 + ab + b2
Certain trinomials that are not quadratic can be expressed as a quadratic trinomials by
making suitable variable substitutions. A trinomial is quadratic in form if it can be
written as au2 + bu + c.
General Factoring Strategy
When factoring a polynomial completely, ask the following questions about the polynomial.
1. Is there a common factor? If so, factor out the GCF.
2. If the polynomial is a binomial, is it the difference of two perfect squares, the sum of
two perfect cubes, or the difference of two perfect cubes? If so, factor.
3. If the polynomial is a trinomial, is it a perfect-square trinomial or the product of two
binomials? If so, factor.
4. Can the polynomial be factored by grouping? If so, factor.
5. Is each factor non-factorable over the integers? If not, factor.
Chapter 11 Section 5: Solving Equations
Principle of Zero Products
If the product of two (or more) factors is zero, then at least one of the factors must be zero.
If a · b = 0 then a = 0 or b = 0. Steps in Solving a Quadratic Equation by Factoring
1. Write the equation in standard form.
2. Factor the polynomial.
3. Set each factor equal to zero.
4. Solve each equation for the variable.
5. Check the solutions.
Chapter 12 Section 1: Multiplication and Division of
Rational Expressions
A rational expression is a fraction in which the numerator and the denominator are
polynomials. A rational expression is in simplest form when the numerator and the denominator have no common factors other than 1. To simplify the rational expression we use
5
A·C
A
the property
= .
B·C
B
Multiply Rational Expressions
To multiply rational expressions we use the property
a c
ac
· = .
b d
bd
Divide Rational Expressions
To divide rational expressions we use the property
a d
a c
÷ = ·
b d
b c
Figure 4: Dividing Rational Expressions Using “Keep Change Flip”
Chapter 12 Section 2: Addition and Subtraction of Rational Expressions
To find the least common multiple (LCM) of two or more polynomials, first factor each
polynomial completely. The LCM is the product of each factor the greatest number of times
it occurs in any one factorization.
Adding and Subtracting Rational Expressions with Common Denominators To
add or subtract rational expressions with the same denominator, we use the properties
a c
a+c
a c
a−c
+ =
or − =
b b
b
b b
b
Adding and Subtraction Rational Expressions without Common Denominators
To add or subtract rational expressions without the same denominator, we find the least
common multiple of the denominators and express each fraction in terms of the common
denominator.
6
Below are some examples for us to try with solutions at the end.
1. Simplify (−3ab)2 (−2ab)3 .
2. Multiply ab (4a2 − 3ab − 8b2 ).
3. Multiply (3x − 7y) (3x + 5y).
4. Multiply (6x − 4y)2 .
5. Simplify
22a4 b8 c4
33a7 b5 c
6. Write the number 0.0000097 in scientific notation.
7. Write the number 7, 320, 000 in scientific notation.
8. Divide
24a2 b + 3ab − 21ab2
.
3ab
9. Divide (2x2 + 15x + 7) ÷ (x + 7).
10. Factor 4a3 + 122 + 8a.
11. Factor 3p3 − 16p2 + 5p by grouping.
12. Factor 1 − 64b3 .
13. Factor 2x4 − 13x2 − 15.
14. Factor 7x2 − 567.
15. Solve the equation z 2 + 5z − 14 = 0.
y 2 + y − 20 y 2 + 5y − 24
16. Multiply 2
·
.
y + 2y − 15 y 2 + 4y − 32
17. Divide
6a2 y + 3a2
12ay + 6a
÷ 3
.
3
2
2x + 4x
6x + 12x2
18. Simplify
2x − 4 x + 1
−
.
4y 2
6xy
7
Solutions
Solution to 1:
(−3ab)2 (−2ab)3 = (−3)2 a2 b2 (−2)3 a3 b3
= 9a2 b2 · (−8)a3 b3
= −72a2+3 b2+3
= −72a5 b5
Solution to 2:
ab(4a2 − 3ab − 8b2 ) = ab(4a2 ) + ab(−3ab) + ab(−8b2 )
= 4a2+1 b − 3a1+1 b1+1 − 8ab1+2
= 4a3 b − 3a2 b2 − 8ab3
Solution to 3:
(3x − 7y) (3x + 5y) = (3x)(3x) + (3x)(5y) + (−7y)(3x) + (−7y)(5y)
= 9x2 + 15xy − 21xy − 35y 2
= 9x2 − 6xy − 35y 2
Solution to 4:
(6x − 4y)2 = (6x − 4y) (6x − 4y)
= (6x)(6x) + (6x)(−4y) + (−4y)(6x) + (−4y)(−4y)
= 36x2 − 24xy − 24xy + 16y 2
= 36x2 − 48xy + 16y 2
Solution to 5:
22a4 b8 c4
2 · 11b8−5 c4−1
=
33a7 b5 c
3 · 11a7−4
2b3 c3
=
3a3
Solution to 6:
To write 0.0000097 in scientific notation, we will move the decimal point to the right until
there is only one nonzero number to the left of the decimal point. Hence, we will move the
decimal point to the right 6 places. Since we are moving the decimal point to the right, the
exponent of 10 will be negative.
0.0000097 = 9.7 × 10−6
8
Solution to 7:
To write 7, 320, 000 in scientific notation, first note that 7, 320, 000 = 7, 320, 000.0. We will
move the decimal point to the left until there is only one nonzero number to the left of the
decimal point. Hence, we will move the decimal point to the left 6 places. Since we are
moving the decimal point to the left, the exponent of 10 will be positive.
7, 320, 000 = 7.32 × 106
Solution to 8: We will use the laws of exponents along with the property
a+b
a b
= + .
c
c c
24a2 b + 3ab − 21ab2
24a2 b 3ab 21ab2
=
+
−
3ab
3ab
3ab
3ab
= 8a2−1 + 1 − 7b2−1
= 8a + 1 − 7b
Solution to 9:
2x + 1Hence, the solution is 2x + 1.
x+7
2x + 15x + 7
− 2x2 − 14x
x+7
−x−7
0
2
Solution to 10:
4a3 + 12a2 + 8a = 4a(a2 + 3a + 2)
= 4a(a + 1)(a + 2)
To factor a2 + 3a + 2, we ask ourselves “What two numbers multiply to give 2 and add to
give 1?” Note that 2 · 1 = 2 AND 2 + 1 = 3.
Solution to 11:
3p3 − 16p2 + 5p = p(3p2 − 16p + 5)
= p(3p2 − 15p + −p + 5)
= p (3p2 − 15p) + (−p + 5)
= p [3p(p − 5) + −1(p − 5)]
= p (p − 5) (3p − 1)
Solution to 12:
Note: 13 = 1 and 64b3 = (4b)3 . Hence, this polynomial is a difference of cubes. Thus, we
can use the difference of cubes formula a3 − b3 = (a − b)(a2 + ab + b2 ) with a = 1 and b = 4b.
9
1 − 64b3 = 13 − (4b)3
= (1 − 4b)(12 + 1(4b) + (4b)2 )
= (1 − 4b)(1 + 4b + 16b2 )
= (1 − 4b)(16b2 + 4b + 1)
Solution to 13:
2
This polynomial is quadratic in form. By letting u = x2 , u2 = (x2 ) = x4 , the polynomial
2x4 − 13x2 − 15 becomes 2u2 − 13u − 15. Once we have factored this polynomial, we will
replace u with x2 .
2x4 − 13x2 − 15 = 2u2 − 13u − 15
= 2u2 − 15u + 2u − 15
= (2u2 − 15u) + (2u − 15)
= u(2u − 15) + 1(2u − 15)
= (u + 1)(2u − 15)
= (x2 + 1)(2x2 − 15)
Solution to 14:
We will use the difference of squares formula a2 − b2 = (a − b)(a + b).
7x2 − 567 = 7(x2 − 81)
= 7(x2 − 92 )
= 7(x − 9)(x + 9)
Solution to 15:
To solve this equation, we will factor the left hand side and then use the zero product property. To factor the left hand side, we ask ourselves “What two numbers multiply to give -14
and add to give 5?” Note that 7 · (−2) = −14 and 7 + (−2) = 5.
z 2 + 5z − 14 = 0
(z + 7)(z − 2) = 0
z + 7 = 0z − 2 = 0
z = −7z = 2
Solution to 16:
To multiply two rational expressions, we factor each numerator and denominator and use
A·C
A
= .
the property
B·C
B
10
y 2 + y − 20 y 2 + 5y − 24
(y + 5)(y − 4) (y + 8)(y − 3)
· 2
=
·
2
y + 2y − 15 y + 4y − 32
(y + 5)(y − 3) (y + 8)(y − 4)
(y + 5)(y − 4)(y + 8)(y − 3)
(y + 5)(y − 3)(y + 8)(y − 4)
=1
=
Solution to 17:
6a2 y + 3a2
12ay + 6a
6a2 y + 3a2 6x3 + 12x2
÷
=
·
2x3 + 4x2
6x3 + 12x2
2x3 + 4x2 12ay + 6a
3a2 (2y + 1) 6x2 (x + 2)
=
·
2x2 (x + 2) 6a(2y + 1)
3a2 · 6x2 (2y + 1)(x + 2)
=
6a · 2x2 (x + 2)(2y + 1)
3a2 · 6x2
=
6a · 2x2
3a2
=
2a
3a2−1
=
2
3a
=
2
Solution to 18:
First, we must find the LCM of 4y 2 and 6xy. To find the LCM, we use each variable with
the largest exponent and the largest product of each prime number. Hence, the LCM of 4y 2
and 6xy is 2 · 2 · 3 · x · y 2 = 12xy 2 Note that 4y 2 · 3x = 12xy 2 and 6xy · 2y = 12xy 2 .
2x − 4 x + 1
2x − 4 3x x + 1 2y
−
=
−
2
4y
6xy
4y 2 3x
6xy 2y
3x(2x − 4) 2y(x + 1)
=
−
12xy 2
12xy 2
2
6x − 12x 2xy + 2y
=
−
12xy 2
12xy 2
2
6x − 12x − 2xy − 2y
=
12xy 2
2
3x − 6x − xy − y
=
6xy 2
11