Transcript A x

Unit 6 – Chapter 9
Unit 6
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Chapter 8 Review and Chap. 8 Skills
Section 9.1 – Adding and Subtracting Polynomials
Section 9.2 – Multiply Polynomials
Section 9.3 – Special Products of Polynomials
Section 9.4 – Solve Polynomial Equations
Section 9.5 – Factor x2 + bx + c
Section 9.6 - Factor ax2 + bx + c
Section 9.7 and 9.8 – Factoring Special Products
and Factoring Polynomials Completely
Warm-Up – X.X
Vocabulary – X.X
• Holder
• Holder 2
• Holder 3
• Holder 4
Notes – X.X – LESSON TITLE.
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Examples X.X
Warm-Up – Chapter 9
Prerequisite Skills
SKILL CHECK
Simplify the expression.
7.
3x +(– 6x)
ANSWER
–3 x
8.
5 + 4x + 2
ANSWER
4x + 7
9.
4(2x – 1) + x
ANSWER
9x – 4
10.
– (x + 4) – 6 x
ANSWER
– 7x – 4
Prerequisite Skills
SKILL CHECK
Simplify the expression.
11. (3xy)3
ANSWER
27x3y3
12.
xy2 xy3
ANSWER
x2 y5
13.
(x5)3
ANSWER
x15
ANSWER
–x3
14. (–
x)3
Vocabulary – 9.1
• Monomial
• Number, variable, or
product of them
• Degree of a Monomial
• Sum of the exponents of
the variables in a term
• Polynomial
• Monomial or Sum of
monomials with
multiple terms
• Degree of a polynomial
• Term with highest
degree
• Leading Coefficient
• Coefficient of the
highest degree term
• Binomial
• Polynomial with 2 terms
• Trinomial
• Polynomial with 3 terms
Notes – 9.1 - Polynomials
• CLASSIFYING POLYNOMIALS
•What is NOT a polynomial? Terms with:
1. Negative exponents
2. Fractional exponents
3. Variables as exponents
•EVERYTHING ELSE IS A POLYNOMIAL!
•To find DEGREE of a term
•Add the exponents of each variable
•Mathlish Polynomial Grammar
•All polynomials are written so that the degree of
the exponents decreases (i.e. biggest first)
Notes – 9.1 – Polynomials – Cont.
• I can only combine things in math that ……????
•ADDING POLYNOMIALS
1. Combine like terms
2. Remember to include the signs of the
coefficients!
•SUBTRACTING POLYNOMIALS
1. Use distributive property first!!
2. Combine like terms
3. Remember to include the signs of the
coefficients!
Examples 9.1
EXAMPLE 1
Rewrite a polynomial
Write 15x – x3 + 3 so that the exponents decrease from
left to right. Identify the degree and leading coefficient
of the polynomial.
SOLUTION
Consider the degree of each of the polynomial’s terms.
15x – x3 + 3
The polynomial can be written as – x3 +15x + 3. The
greatest degree is 3, so the degree of the polynomial is
3, and the leading coefficient is –1.
EXAMPLE 2
Identify and classify polynomials
Tell whether is a polynomial. If it is a polynomial, find
its degree and classify it by the number of its terms.
Otherwise, tell why it is not a polynomial.
Expression Is it a polynomial?
Classify by degree and
number of terms
a.
9
Yes
0 degree monomial
b.
c.
d.
e.
2x2 + x – 5
Yes
2nd degree trinomial
6n4 – 8n
No; variable exponent
n– 2 – 3
No; variable exponent
7bc3 + 4b4c
Yes
5th degree binomial
EXAMPLE 3
Add polynomials
Find the sum.
a. (2x3 – 5x2 + x) + (2x2 + x3 – 1)
b.
(3x2 + x – 6) + (x2 + 4x + 10)
EXAMPLE 3
Add polynomials
SOLUTION
a.
Vertical format: Align like
terms in vertical columns.
(2x3 – 5x2 + x)
+ x3 + 2x2
–1
3x3 – 3x2 + x – 1
b.
Horizontal format: Group like terms and simplify.
(3x2 + x – 6) + (x2 + 4x + 10) = (3x2 + x2) + (x + 4x) + (– 6 + 10)
= 4x2 + 5x + 4
EXAMPLE
1
Examples 1,2, and 3
Rewrite afor
polynomial
GUIDED PRACTICE
1.
Write 5y – 2y2 + 9 so that the exponents decrease from
left to right. Identify the degree and leading coefficient
of the polynomial.
SOLUTION
Consider the degree of each of the polynomial’s terms.
Degree is 1
Degree is 2
Degree is 0
5y –2y2 + 9
The polynomial can be written as – 2y2 +5y + 9. The
greatest degree is 2, so the degree of the polynomial is
2, and the leading coefficient is –2
EXAMPLE
2
for classify
Examplepolynomials
Examples
1,2, and 3
Identify and
GUIDED PRACTICE
2.
Tell whether y3 – 4y + 3 is a polynomial. If it is a
polynomial, find its degree and classify it by the
number of its terms. Otherwise, tell why it is not a
polynomial.
SOLUTION
y3 – 4y + 3 is a polynomial. 3 degree trinomial.
EXAMPLE
3
for Examples
Example 1,2, and 3
Add polynomials
GUIDED PRACTICE
3.
Find the sum.
a.
(2x3 + 4x – x) + (4x2 +3x3 – 6)
EXAMPLE
3
for Examples
Example 1,2, and 3
Add polynomials
GUIDED PRACTICE
SOLUTION
a.
Vertical format: Align like
terms in vertical columns.
(5x3 + 4x – 2x)
+ 3x3 + 4x2
–6
8x3 + 4x2 +2x – 6
b.
Horizontal format: Group like terms and simplify.
(5x3 +4x – 2) + (4x2 + 3x3 – 6) = (5x3 + 3x3) + (4x2) + (4x –2x) + (– 6)
= 8x3 + 4x2 + 2x – 6
EXAMPLE 4
Subtract polynomials
Find the difference.
a. (4n2 + 5) – (– 2n2 + 2n – 4)
b.
(4x2 – 3x + 5) – (3x2 – x – 8)
EXAMPLE 4
Subtract polynomials
SOLUTION
a.
(4n2
+ 5)
– (– 2n2 + 2n – 4)
4n2
+5
2n2 – 2n + 4
6n2 – 2n + 9
b. (4x2 – 3x + 5) – (3x2 – x – 8) = 4x2 – 3x + 5 – 3x2 + x + 8
= (4x2 – 3x2) + (– 3x + x) + (5 + 8)
= x2 – 2x + 13
EXAMPLE
4
for Examples 4 and 5
Subtract polynomials
GUIDED PRACTICE
4.
Find the difference.
a. (4x2 + 7x) – ( 5x2 + 4x – 9)
(4x2 – 7x ) – (5x2 – 4x – 9) = 4x2 – 7x – 5x2 + 4x + 9
= (4x2 – 5x2) + (– 7x – 4x) + 9
= –x2 – 11x + 9
EXAMPLE 5
Solve a multi-step problem
5. BASEBALL ATTENDNCE Look back at Example 5. Find
The difference in attendance at National and American
League baseball games in 2001.
M = (– 488t2 + 5430t + 24,700) – (– 318t2 + 3040t + 25,600)
= – 488t2 + 5430t + 24,700 + 318t2 – 3040t – 25,600
= (– 488t2 + 318t2) + (5430t – 3040t) + (24,700 – 25,600)
= – 170t2 + 2390t – 900
= 7320
Substitute 6 for t in the model, because 2001 is
6 years after 1995.
EXAMPLE 5
Solve a multi-step problem
ANSWER
About 7,320,000 people attended Major League
Baseball games in 2001.
Warm-Up – 9.2
Lesson 9.2, For use with pages 561-568
1. Simplify –2 (9a – b).
4. Simplify x2(x+1) + 2x(3x+3) + 2x +5
ANSWER
–18a + 2b
ANSWER
2. Simplify r2s rs3.
ANSWER
r 3 s4
3. Simplify 2x(3x + 2)
ANSWER
6x2 + 4x
x3 + 7x2 + 8x + 5
Lesson 9.2, For use with pages 561-568
3. The number of hardback h and paperback p books
(in hundreds) sold from 1999–2005 can be modeled by
h = 0.2t2 – 1.7t + 14 and p = 0.17t3 – 2.7t2 + 11.7t + 27 where
t is the number of years since 1999. About how many
books sold in 2003.
ANSWER
5200
4. Simplify (x + 1)(x + 2)
ANSWER
x2 + 3x +2
Vocabulary – 9.2
• Polynomial
• Monomial or Sum of
monomials with multiple
terms
Notes – 9.2 – Multiply Polynomials
• Multiplying Polynomials is like using the distributiv
property over and over and over again.
•Everything must be multiplied by everything else
and combine like terms!!!
• Frequently people use the FOIL process to multiply
polynomials.
• F – Multiply the First Terms
•O – Multiply the Outside Terms
•I – Multiply the Inside Terms
•L – Multiply the Last Terms
Examples 9.2
EXAMPLE 1
Multiply a monomial and a polynomial
Find the product 2x3(x3 + 3x2 – 2x + 5).
2x3(x3 + 3x2 – 2x + 5)
Write product.
= 2x3(x3) + 2x3(3x2) – 2x3(2x) + 2x3(5)
Distributive property
= 2x6 + 6x5 – 4x4 + 10x3
Product of powers property
EXAMPLE 2
Multiply polynomials using a table
Find the product (x – 4)(3x + 2).
SOLUTION
STEP 1
Write subtraction as addition in each polynomial.
(x – 4)(3x + 2) = [x + (– 4)](3x + 2)
EXAMPLE 2
Multiply polynomials using a table
STEP 2
Make a table of products.
3x
x
–4
3x2
2
x
3x
2
3x2
2x
– 4 – 12x
ANSWER
The product is 3x2 + 2x – 12x – 8, or 3x2 – 10x – 8.
–8
GUIDED PRACTICE
for Examples 1 and 2
Find the product.
1 x(7x2 +4)
SOLUTION
x(7x2 +4)
= x(7x2 )+x(4)
= 7x3+4x
Write product.
Distributive property
Product of powers property
GUIDED PRACTICE
for Examples 1 and 2
Find the product.
2 (a +3)(2a +1)
SOLUTION
Make a table of products.
2a
1
a
2a2
a
3
6a
3
ANSWER
The product is 2a2 + a + 6a + 3, or 2a2 + 7a + 3.
GUIDED PRACTICE
for Examples 1 and 2
Find the product.
3 (4n – 1) (n +5)
SOLUTION
STEP 1
Write subtraction as addition in each polynomial.
(4n – 1) (n +5) = [4n + (– 1)](n +5)
for Examples 1 and 2
GUIDED PRACTICE
STEP 2
Make a table of products.
n
5
4n
4n2
20n
–1
–n
–5
ANSWER
The product is 4n2 + 20n – n – 5, or 4n2 + 19n – 5.
EXAMPLE 3
Multiply polynomials vertically
Find the product (b2 + 6b – 7)(3b – 4).
SOLUTION
3b3 + 14b2 – 45b + 28
EXAMPLE 4
Multiply polynomials horizontally
Find the product (2x2 + 5x – 1)(4x – 3).
Solution: Multiply everything and get
= (2x2)(4x) + (2x2)(-3) + (5x)(4x) + (5x)(-3) + (-1)(4x) + (-1)(-3)
= 8x3 + 14x2 – 19x + 3
FOIL PATTERN The letters of the word FOIL can help
you to remember how to use the distributive property to
multiply binomials. The letters should remind you of the
words First, Outer, Inner, and Last.
First Outer Inner Last
(2x + 3)(4x + 1) = 8x2 + 2x + 12x + 3
EXAMPLE 5
Multiply binomials using the FOIL pattern
Find the product (3a + 4)(a – 2).
(3a + 4)(a – 2)
= (3a)(a) + (3a)(– 2) + (4)(a) + (4)(– 2) Write products of terms.
= 3a2 + (– 6a) + 4a + (– 8)
Multiply.
= 3a2 – 2a – 8
Combine like terms.
for Examples 3, 4, and 5
GUIDED PRACTICE
Find the product.
4 (x2 + 2x +1)(x + 2)
SOLUTION
x3 + 4x2 + 5x + 2
GUIDED PRACTICE
for Examples 3, 4, and 5
Find the product.
5 (3y2 –y + 5)(2y – 3)
SOLUTION
(3y2 –y + 5)(2y – 3)
Write product.
= 3y2(2y – 3) – y(2y – 3) + 5(2y – 3)
Distributive property
= 6y3 – 9y2 – 2y2 + 3y + 10y – 15
Distributive property
= 6y3 – 11y2 + 13y – 15
Combine like terms.
GUIDED PRACTICE
for Examples 3, 4, and 5
Find the product.
6 (4b –5)(b – 2)
SOLUTION
= (4b)(b) + (4b)(– 2) + (–5)(b) + (–5)(– 2) Write products of terms.
= 4b2 – 8b – 5b + 10
Multiply.
= 4b2 – 13b + 10
Combine like terms.
EXAMPLE 6
Standardized Test Practice
The dimensions of a rectangle are x + 3 and x + 2.
Which expression represents the area of the
rectangle?
A
x2 + 6
B
x2 + 5x + 6
C
x2 + 6x + 6
D
x2 + 6x
SOLUTION
Area = length width
Formula for area of a rectangle
= (x + 3)(x + 2)
Substitute for length and width.
= x2 + 2x + 3x + 6
Multiply binomials.
EXAMPLE 6
Standardized Test Practice
= x2 + 5x + 6
Combine like terms.
ANSWER
The correct answer is B.
A
B
CHECK
You can use a graph to check your
answer. Use a graphing calculator to
display the graphs of y1 = (x + 3)(x + 2)
and y2 = x2 + 5x + 6 in the same
viewing window. Because the graphs
coincide, you know that the product
of x + 3 and x + 2 is x2 + 5x + 6.
C
D
Warm-Up – 9.3
Lesson 9.3, For use with pages 569-574
Find the product.
1. (x + 7)(x + 7)
ANSWER
x2 + 14x + 49
3. (x + 7)(x - 7)
ANSWER
x2 – 49
2. (x - 7)(x - 7)
2. (3x – 1)(3x + 2)
ANSWER
x2 - 14x + 49
ANSWER
9x2 + 3x – 2
Lesson 9.3, For use with pages 569-574
Find the product.
3. The dimensions of a rectangular playground can be
represented by 3x + 8 and 5x + 2. Write a polynomial
that represents the area of the playground. What is
the area of the playground if x is 8 meters?
ANSWER
15x2 + 46x + 16; 1344 m2
Vocabulary – 9.3
• Binomial
• Polynomial with 2 terms
• Trinomial
• Polynomial with 3 terms
Notes –9.3 – Special Products of Poly.
• Find the area of the larger square
•Multiply (a+b)2
• = (a+b)(a+b)
• = a2 + 2ab + b2
•Multiply (a – b) 2
• = (a – b)(a – b) = a2 - 2ab + b2
Notes –9.3 – Special Products of Poly.
• Multiply (a + b)(a - b)
= a2 - b 2
•This is a very special type of polynomial called the
DIFFERENCE OF TWO SQUARES
Examples 9.4
EXAMPLE 1
Use the square of a binomial pattern
Find the product.
a.
(3x + 4)2 =(3x)2 + 2(3x)(4) + 42
= 9x2 + 24x + 16
b.
(5x – 2y)2 = (5x)2 – 2(5x)(2y) + (2y)2
=25x2 – 20xy + 4y2
Square of a binomial
pattern
Simplify.
Square of a binomial
pattern
Simplify.
GUIDED PRACTICE
for Example 1
Find the product.
1.
(x + 3)2 = (x)2 + 2(x)(3) + (3)2
= x2 + 6x + 9
2.
(2x + 1)2 = (2x)2 + 2(2x)(1) + (1)2
=4x2 + 4x + 1
Square of a binomial
pattern
Simplify.
Square of a binomial
pattern
Simplify.
GUIDED PRACTICE
3.
for Example 1
(4x – y)2 = (4x)2 – 2(4x)(y) + (y)2
= 16x2 – 8xy + y2
4.
(3m +n)2 = (3m)2 + 2(3m)(n) + (n)2
= 9m2 + 6mn + n2
Square of a binomial
pattern
Simplify.
Square of a binomial
pattern
Simplify.
EXAMPLE 2
Use the sum and difference pattern
Find the product.
a.
b.
(t + 5)(t – 5)
= t2 – 52
Sum and difference pattern
= t2 – 25
Simplify.
(3x + y)(3x – y) = (3x)2 – y2
= 9x2 – y2
Sum and difference pattern
Simplify.
GUIDED PRACTICE
for Example 2
Find the product.
5.
(x + 10)(x – 10) = x2 – 102
= x2 – 100
6. (2x + 1)(2x – 1) = (2x)2 – 12
= 4x2 – 1
7.
(x + 3y)(x – 3y) = (x)2 – (3y)2
= x2 – 9y2
Sum and difference pattern
Simplify.
Sum and difference pattern
Simplify.
Sum and difference pattern
Simplify.
EXAMPLE 3
Use special products and mental math
Use special products to find the product 26 34.
SOLUTION
Notice that 26 is 4 less than 30 while 34 is 4 more than 30.
26 34 = (30 – 4)(30 + 4) Write as product of difference and sum.
= 302 – 42
= 900 – 16
Sum and difference pattern
= 884
Simplify.
Evaluate powers.
EXAMPLE 4
Solve a multi-step problem
Border Collies
The color of the dark patches of a border collie’s coat
is determined by a combination of two genes. An
offspring inherits one patch color gene from each
parent. Each parent has two color genes, and the
offspring has an equal chance of inheriting either one.
EXAMPLE 4
Solve a multi-step problem
The gene B is for black patches,
and the gene r is for red patches.
Any gene combination with a B
results in black patches. Suppose
each parent has the same gene
combination Br. The Punnett
square shows the possible gene
combinations of the offspring and
the resulting patch color.
What percent of the possible gene combinations
of the offspring result in black patches?
Show how you could use a polynomial to model
the possible gene combinations of the offspring.
EXAMPLE 4
Solve a multi-step problem
SOLUTION
STEP 1
Notice that the Punnett square shows 4 possible
gene combinations of the offspring. Of these
combinations, 3 result in black patches.
ANSWER
75% of the possible gene combinations result in black
patches.
EXAMPLE 4
Solve a multi-step problem
STEP 2
Model the gene from each parent with 0.5B + 0.5r.
There is an equal chance that the collie inherits a
black or red gene from each parent.
The possible genes of the offspring can be
modeled by (0.5B + 0.5r)2. Notice that this product
also represents the area of the Punnett square.
Expand the product to find the possible patch
colors of the offspring.
(0.5B + 0.5r)2 =(0.5B)2 + 2(0.5B)(0.5r) + (0.5r)2
= 0.25B2 + 0.5Br + 0.25r2
EXAMPLE 4
Solve a multi-step problem
Consider the coefficients in the polynomial.
= 0.25B2 + 0.5Br + 0.25r2
The coefficients show that 25% + 50% = 75% of
the possible gene combinations will result in
black patches.
GUIDED PRACTICE
8.
for Examples 3 and 4
Describe how you can use special product to find 212.
Use the square of binomial pattern to find the product
(20 +1)2.
(20 + 1)2 = (20)2 + 2(20)(1) + 12
= 421
Square of a binomial
pattern
Simplify.
GUIDED PRACTICE
for Examples 3 and 4
BORDER COLLIES
Look back at Example 4. What percent of the possible
gene combinations of the offspring result in red
patches?
SOLUTION
STEP 1
Notice that the Punnett square shows 4 possible
gene combinations of the offspring. Of these
combinations, 1 result in red patches.
ANSWER
25% of the possible gene combinations result in red
patches.
GUIDED PRACTICE
for Examples 3 and 4
STEP 2
Model the gene from each parent with 0.5B + 0.5r.
There is an equal chance that the collie inherits a
black or red gene from each parent.
The possible genes of the offspring can be
modeled by (0.5B + 0.5r)2. Notice that this product
also represents the area of the Punnett square.
Expand the product to find the possible patch
colors of the offspring.
(0.5B + 0.5r)2 = (0.5B)2 + 2(0.5B)(0.5r) + (0.5r)2
= 0.25B2 + 0.5Br + 0.25r2
GUIDED PRACTICE
for Examples 3 and 4
Consider the coefficients in the polynomial.
= 0.25B2 + 0.5Br + 0.25r2
The coefficients show that 25% of the possible
gene combinations will result in red patches.
Warm-Up – 9.4
Lesson 9.4, For use with pages 575-580
1. Find the GCF of 12 and 28.
ANSWER
4
3. Find the binomials that
multiply to get x2 - 9
2. (2a – 5b)(2a + 5b)
ANSWER
4a2 – 25b2
ANSWER
(x + 3)(x - 3)
Lesson 9.4, For use with pages 575-580
3. The number (in hundreds) of sunscreen and sun
tanning products sold at a pharmacy from 1999–2005
can be modeled by –0.8t2 + 0.3t + 107, where t is the
number of years since 1999. About how many
products were sold in 2002?
ANSWER
about 10,070
1. Write down an equation with values for X and Y that
make the following equation true:
X*Y=0
Vocabulary – 9.4
• Roots
• Solutions for x when a function = 0
• Also where the function crosses the X axis
Notes – 9.4 – Solve Polynomial Eqns.
SOLVING EQUATIONS THAT EQUAL ZERO
• Use the Zero Product Property to solve algebraic
equations that equal 0.
• Solutions to algebraic equations that equal zero are called
the “roots” or the “zeros” of a function.
FACTORING
• To solve a polynomial equation, it may be necessary to
“break it up” into its factors.
• Find the GCF of ALL the terms and factor it out. It’s like
“unmultiplying” the distributive property.
Notes – 9.4 – Solve Polynomial Eqns.
VERTICAL MOTION MODEL
• The height of an object (in FEET!!) can be modeled by the
following equation:
•
H(t) = -16t2 + vt + s
s = initial height (in feet)
v = initial velocity (in feet/second)
Height of the object
Above the ground
t = time (in seconds)
Examples 9.4
EXAMPLE 1
Use the zero-product property
Solve (x – 4)(x + 2) = 0.
(x – 4)(x + 2) = 0
x – 4 = 0 or x + 2 = 0
x = 4 or
x=–2
Write original equation.
Zero-product property
Solve for x.
ANSWER
The solutions of the equation are 4 and –2.
GUIDED PRACTICE
for Example 1
1. Solve the equation (x – 5)(x – 1) = 0.
(x – 5)(x – 1) = 0
x – 5 = 0 or x – 1 = 0
x = 5 or
x=1
Write original equation.
Zero-product property
Solve for x.
ANSWER
The solutions of the equation are 5 and 1.
EXAMPLE 2
Find the greatest common monomial factor
Factor out the greatest common monomial factor.
a.
12x + 42y
SOLUTION
a.
The GCF of 12 and 42 is 6. The variables x and y have
no common factor. So, the greatest common monomial
factor of the terms is 6.
ANSWER
12x + 42y = 6(2x + 7y)
EXAMPLE 2
Find the greatest common monomial factor
Factor out the greatest common monomial factor.
b. 4x4 + 24x3
SOLUTION
b.
The GCF of 4 and 24 is 4. The GCF of x4 and x3 is x3. So,
the greatest common monomial factor of the terms is
4x3.
ANSWER
4x4 + 24x3 = 4x3(x + 6)
GUIDED PRACTICE
2.
for Example 2
Factor out the greatest common monomial factor
from 14m + 35n.
SOLUTION
The GCF of 14 and 35 is 7. The variables m and n have no
common factor. So, the greatest common monomial factor of
the terms is 7.
ANSWER
14m + 35n = 7(2m + 5n)
EXAMPLE 3
Solve an equation by factoring
Solve 2x2 + 8x = 0.
2x2 + 8x = 0.
Write original equation.
2x(x + 4) = 0
2x = 0
x=0
Factor left side.
or x + 4 = 0
or
x=–4
Zero-product property
Solve for x.
ANSWER
The solutions of the equation are 0 and – 4.
EXAMPLE 4
Solve an equation by factoring
Solve 6n2 = 15n.
6n2 – 15n = 0
Subtract 15n from each side.
3n(2n – 5) = 0
3n = 0
n=0
Factor left side.
or 2n – 5 = 0
or
n=
5
2
Zero-product property
Solve for n.
ANSWER
5
The solutions of the equation are 0 and .
2
for Examples 3 and 4
GUIDED PRACTICE
Solve the equation.
3. a2 + 5a = 0.
a2 + 5a = 0
Write original equation.
a(a + 5) = 0
Factor left side.
a=0
or a + 5 = 0
a=0
or
a=–5
Zero-product property
Solve for x.
ANSWER
The solutions of the equation are 0 and – 5.
for Examples 3 and 4
GUIDED PRACTICE
4. 3s2 – 9s = 0.
3s2 – 9s = 0
Write original equation.
3s(s – 3) = 0
3s = 0
s= 0
Factor left side.
or s – 3 = 0
Zero-product property
or
Solve for x.
s=3
ANSWER
The solutions of the equation are 0 and 3.
for Examples 3 and 4
GUIDED PRACTICE
5.
4x2 = 2x.
4x2 = 2x
Write original equation.
4x2 – 2x = 0
Subtract 2x from each side.
2x(2x – 1) = 0
2x = 0
x=0
Factor left side.
or 2x – 1 = 0
or
x=
1
2
Zero-product property
Solve for x.
ANSWER
1
The solutions of the equation are 0 and .
2
EXAMPLE 5
Solve a multi-step problem
ARMADILLO
A startled armadillo jumps
straight into the air with an
initial vertical velocity of 14
feet per second.After how
many seconds does it land
on the ground?
EXAMPLE 5
Solve a multi-step problem
SOLUTION
STEP 1
Write a model for the armadillo’s height above the
ground.
h = – 16t2 + vt + s
Vertical motion model
h = – 16t2 + 14t + 0
Substitute 14 for v and 0 for s.
h = – 16t2 + 14t
Simplify.
EXAMPLE 5
Solve a multi-step problem
STEP 2
Substitute 0 for h. When the armadillo lands, its height
above the ground is 0 feet. Solve for t.
0 = – 16t2 + 14t
Substitute 0 for h.
0 = 2t(–8t + 7)
Factor right side.
2t = 0
or
t=0
or
–8t + 7 = 0
t = 0.875
Zero-product property
Solve for t.
ANSWER
The armadillo lands on the ground 0.875 second after
the armadillo jumps.
GUIDED PRACTICE
for Example 5
6. WHAT IF? In Example 5, suppose the initial
vertical velocity is 12 feet per second.After how
many seconds does armadillo land on the
ground?
SOLUTION
STEP 1
Write a model for the armadillo’s height above the
ground.
h = – 16t2 + vt + s
Vertical motion model
h = – 16t2 + 12t + 0
Substitute 12 for v and 0 for s.
h = – 16t2 + 12t
Simplify.
GUIDED PRACTICE
for Example 5
STEP 2
Substitute 0 for h. When the armadillo lands, its height
above the ground is 0 feet. Solve for t.
0 = – 16t2 + 12t
Substitute 0 for h.
0 = – 4t(4t – 3)
Factor right side.
– 4t = 0
or
t=0
or
4t – 3 = 0
t = 0.75
Zero-product property
Solve for t.
ANSWER
The armadillo lands on the ground 0.75 second after
the armadillo jumps.
Warm-Up – 9.5
Lesson 9.5, For use with pages 582-589
Find the product.
1. (y + 3)(y - 5)
3. (y - 3)( y - 5)
ANSWER
y2 - 2x – 15
ANSWER
2. (y + 3)( y + 5)
ANSWER
y2 + 8y + 15
y2 – 8y + 15
4. (2y + 3)( y + 5)
ANSWER
2y2 + 13y + 15
Vocabulary – 9.5
• Zero of a Function
• The X value(s) where a function equals zero
• AKA the “roots” of a function
• Factoring a Polynomial
• Finding the polynomial factors that will multiply to
get the original.
• It’s “unFOILing” or “unmultiplying” a polynomial
Notes – 9.5 – Factor x2+bx+c
• If you multiply two binomials (x + p)(x + q) to get
x2 + bx + c, the following must be true:
•p * q = c
•p + q = b (NOTICE ALL X COEFF. ARE = 1!!)
•We can use these truths to go the other direction, and
factor a polynomial into binomials
(x + p)(x + q)
Polynomial
x2 + bx + c
Signs of b and c
(x + 2)(x + 3)
x2 + 5x + 6
b and c positive
(x + 2)(x - 3)
x2 - x – 6
b and c negative
(x - 2)(x + 3)
x2 + x – 6
b is pos., c is neg.
(x - 2)(x - 3)
x2 - 5x + 6
B is neg., c is pos.
Examples 9.5
EXAMPLE 1
Factor when b and c are positive
Factor x2 + 11x + 18.
SOLUTION
Find two positive factors of 18 whose sum is 11.
Make an organized list.
Factors of 18
Sum of factors
18, 1
18 + 1 = 19
9, 2
9 + 2 = 11
6, 3
6+3=9
Correct sum
EXAMPLE 1
Factor when b and c are positive
The factors 9 and 2 have a sum of 11, so they are the
correct values of p and q.
ANSWER
x2 + 11x + 18 = (x + 9)(x + 2)
CHECK
(x + 9)(x + 2) = x2 + 2x + 9x + 18 Multiply binomials.
= x2 + 11x + 18
Simplify.
GUIDED PRACTICE
for Example 1
Factor the trinomial
1. x2 + 3x + 2.
SOLUTION
Find two positive factors of 2 whose sum is 3.
Make an organized list.
Factors of 2
1, 2
Sum of factors
1+2=3
Correct sum
EXAMPLE
1
for Example
1 positive
Factor when
b and c are
GUIDED PRACTICE
The factors 2 and 1 have a sum of 3, so they are the
correct values of p and q.
ANSWER
x2 + 3x + 2 = (x + 2)(x + 1)
GUIDED PRACTICE
for Example 1
Factor the trinomial
2. a2 + 7a + 10.
SOLUTION
Find two positive factors of 10 whose sum is 7.
Make an organized list.
Factors of 10
Sum of factors
10, 1
10 + 1 = 11
2, 5
2+5=7
Correct sum
EXAMPLE
1
for Example
1 positive
Factor when
b and c are
GUIDED PRACTICE
The factors 5 and 2 have a sum of 7, so they are the
correct values of p and q.
ANSWER
a2 + 7a + 10 = (a + 5)(a + 2)
GUIDED PRACTICE
for Example 1
Factor the trinomial
3. t2 + 9t + 14.
SOLUTION
Find two positive factors of 14 whose sum is 9.
Make an organized list.
Factors of 14
Sum of factors
14, 1
14 + 1 = 15
7, 2
7+2=9
Correct sum
EXAMPLE
1
for Example
1 positive
Factor when
b and c are
GUIDED PRACTICE
The factors 7 and 2 have a sum of 9, so they are the
correct values of p and q.
ANSWER
t2 + 9t + 14 = (t + 7)(t + 2)
EXAMPLE 2
Factor when b is negative and c is positive
Factor n2 – 6n + 8.
Because b is negative and c is positive, p and q must
both be negative.
Factors of 8
Sum of factors
–8,–1
–8 + (–1) = –9
–4,–2
–4 + (–2) = –6
ANSWER
n2 – 6n + 8 = (n – 4)( n – 2)
Correct sum
EXAMPLE 3
Factor when b is positive and c is negative
Factor y2 + 2y – 15.
Because c is negative, p and q must have different
signs.
Factors of –15
–15, 1
15, –1
–5, 3
5, –3
ANSWER
Sum of factors
–15 + 1 = –14
15 + (–1) = 14
–5 + 3 = –2
5 + (–3) = 2
y2 + 2y – 15 = (y + 5)( y – 3)
Correct sum
GUIDED PRACTICE
for Examples 2 and 3
Factor the trinomial
4. x2 – 4x + 3.
Because b is negative and c is positive, p and q must
both be negative.
Factors of 2
–3, –1
Sum of factors
–3 + (–1) = –4
ANSWER
x2 – 4x + 3 = (x – 3)( x – 1)
Correct sum
GUIDED PRACTICE
for Examples 2 and 3
Factor the trinomial
5. t2 – 8t + 12.
Because b is negative and c is positive, p and q must
both be negative.
Factors of 12
Sum of factors
–12, –1
–12 + (–1) = –13
–6, –2
–6 + (–2) = –8
–4, –3
–4 + (–3) = –7
ANSWER
t2 – 8t + 12 = (t – 6)( t – 2)
Correct sum
GUIDED PRACTICE
for Examples 2 and 3
Factor the trinomial
6. m2 + m – 20.
Because c is negative, p and q must have different signs.
Factors of 20
–20, 1
–1, 20
–1 + 20 = 19
–10, 2
– 10 + 2 = –8
–2,10
– 2 +10 = 8
–5, 4
5, –4
ANSWER
Sum of factors
–20 + 1 = –19
–5 + 4 = –1
5–4=1
m2 + m – 20 = (m + 5)( m – 4)
Correct sum
GUIDED PRACTICE
for Examples 2 and 3
Factor the trinomial
7. w2 + 6w – 16.
Because c is negative, p and q must have different signs.
Factors of 16
–16, 1
Sum of factors
–16 + 1 = –15
16, –1
–8, 2
8, –2
16 + (–1) = 15
–4, 4
ANSWER
– 8 + 2 = –6
8 + (–2) = 6
–4 + 4 = 0
w2 + 6w – 16 = (w + 8)( w – 2)
Correct sum
EXAMPLE 4
Solve a polynomial equation
Solve the equation x2 + 3x = 18.
x2 + 3x = 18
x2 + 3x – 18 = 0
(x + 6)(x – 3) = 0
or
x–3 =0
x = – 6 or
x=3
x +6=0
ANSWER
Write original equation.
Subtract 18 from each side.
Factor left side.
Zero-product property
Solve for x.
The solutions of the equation are – 6 and 3.
EXAMPLE
4 Solve a polynomial
for Example
4
equation
GUIDED PRACTICE
8.
Solve the equation s2 – 2s = 24.
s2 – 2s = 24.
s2 – 2s – 24 = 0
(s + 4)(s – 6) = 0
or
s–6 =0
s = – 4 or
s=6
s +4=0
ANSWER
Write original equation.
Subtract 24 from each side.
Factor left side.
Zero product property
Solve for x.
The solutions of the equation are – 4 and 6.
EXAMPLE 5
Solve a multi-step problem
Banner Dimensions
You are making banners to hang
during school spirit week. Each
banner requires 16.5 square feet
of felt and will be cut as shown.
Find the width of one banner.
SOLUTION
STEP 1
Draw a diagram of two banners together.
EXAMPLE 5
Solve a polynomial equation
STEP 2
Write an equation using the fact that the area of 2 banners
is 2(16.5) = 33 square feet. Solve the equation for w.
A=l w
Formula for area of a rectangle
33 = (4 + w + 4) w
Substitute 33 for A and (4 + w + 4) for l.
0 = w2 + 8w – 33
Simplify and subtract 33 from each side.
0 = (w + 11)(w – 3) Factor right side.
w + 11 = 0 or w – 3 = 0 Zero-product property
w = – 11 or w = 3 Solve for w.
The banner cannot have a negative width,
ANSWER
so the width is 3 feet.
GUIDED PRACTICE
for Example 5
What if? In example 5, suppose the area of a banner is
to be 10 square feet. What is the width of one banner?
9.
Write an equation using the fact that the area of 2 banners
is 2(10) = 20 square feet. Solve the equation for w.
A=l w
20 = (4 + w + 4) w
20 = w2 + 8w
0 = w2 + 8w – 20
0 = (w + 10)(w – 2)
w + 10 = 0 or w – 2 = 0
w = – 10 or w = 2
ANSWER
Formula for area of a rectangle
Substitute 20 for A and (4 + w + 4) for l.
Simplify
Subtract 20 from each side.
Factor right side.
Zero-product property
Solve for w.
The banner cannot have a negative width,
so the width is 2 feet.
Warm-Up – 9.6
Lesson 9.6, For use with pages 592-599
Find the product.
1. (3c + 3)(2c – 3)
ANSWER
6c2 – 3c – 9
2. (2y + 3)(2y + 1)
ANSWER
4y2 + 8y + 3
Factor and solve
the polynomials.
3. x2 – x – 6 = 0
ANSWER
(x-3)(x+2)
x = 3 or x= -2
4. x2 + 13x = -36
ANSWER
(x + 9)(x + 4)
x = -4 or x= -9
Lesson 9.6, For use with pages 592-599
Find the product.
3. A cat leaps into the air with an initial velocity of
12 feet per second to catch a speck of dust, and then
falls back to the floor. How long does the cat remain
in the air? H(t) = -16t2 + vt + s
ANSWER
0.75 sec
Vocabulary – 9.6
• Trinomial
• Polynomial with 3 terms
Notes – 9.6 – Factor ax2+bx+c
• If you multiply two binomials (dx + p)(ex + q) to
get ax2 + bx + c, the following must be true:
•d * e = a (NOTICE X COEFF. DO NOT = 1!!)
•p * q = c
•TO FACTOR POLYNOMIALS WHERE A ≠ 1
•If A = -1, factor out -1 from the polynomial
and factor
•If A > 0, use a table to organize your work
Factors of
“a”
Factors of
“c”
Possible
Factorizations
Middle Term
when
multiplied
Examples 9.6
EXAMPLE 1
Factor when b is negative and c is positive
Factor 2x2 – 7x + 3.
SOLUTION
Because b is negative and c is positive, both factors
of c must be negative. Make a table to organize your
work.
You must consider the order of the factors of 3,
because the x-terms of the possible
factorizations are different.
EXAMPLE 1
Factor when b is negative and c is positive
Factor 2x2 – 7x + 3.
Factors Factors
of 2
of 3
Possible
factorization
Middle term
when multiplied
1,2
– 1, – 3
(x – 1)(2x – 3)
– 3x – 2x = – 5x
1, 2
-3, -1
(x – 3)(2x – 1)
– x – 6x = – 7x
ANSWER 2x2 – 7x + 3 =(x – 3)(2x – 1)
Correct
EXAMPLE 2
Factor when b is positive and c is negative
Factor 3n2 + 14n – 5.
SOLUTION
Because b is positive and c is negative, the factors
of c have different signs.
EXAMPLE 2
Factor when b is negative and c is positive
Factors
of 3
Factors of
–5
Possible
factorization
Middle term
when multiplied
1, 3
1, –5
(n + 1)(3n – 5)
– 5n + 3n = – 2n
1, 3
–1, 5
(n – 1)(3n + 5)
5n – 3n = 2n
1, 3
5, – 1
(n + 5)(3n – 1)
– n + 15n = 14n
1, 3
– 5, 1
(n – 5)(3n + 1)
n – 15n = –14n
ANSWER 3n2 + 14n – 5 = (n + 5)(3n – 1)
Correct
GUIDED PRACTICE
for Examples 1 and 2
Factor the trinomial.
1. 3t2 + 8t + 4.
SOLUTION
Because b is positive and c is positive, both factors
of c are positive.
You must consider the order of the factors of 4,
because the t-terms of the possible factorizations
are different.
GUIDED PRACTICE
for Examples 1 and 2
Factors Factors
of 3
of 4
Possible
factorization
Middle term
when multiplied
1, 3
1, 4
(t + 1)(3t + 4)
4t + 3t = 7t
1, 3
4, 1
(t + 4)(3t + 1)
t + 12t = 13t
1, 3
2, 2
(t + 2)(3t + 2)
2t + 6t = 8t
ANSWER
3t2 + 8t + 4 = (t + 2)(3t + 2)
Correct
GUIDED PRACTICE
for Examples 1 and 2
Factor the trinomial.
2.
4s2 – 9s + 5.
SOLUTION
Because b is negative and c is positive, both factors
of c must be negative. Make a table to organize your
work.
You must consider the order of the factors of 5,
because the s-terms of the possible factorizations
are different.
GUIDED PRACTICE
Factors Factors
of 4
of 5
for Examples 1 and 2
Possible
factorization
Middle term
when multiplied
1, 4
– 1, – 5
(s – 1)(4s – 5)
– 5s – 4s = – 9s
1, 4
– 5, – 1
(s – 5)(4s – 1)
– s – 20s = – 21s
2, 2
– 1, – 5
(2s – 1)(2s – 5) – 10s – 2s = – 12s
ANSWER
4s2 – 9s + 5 = (s – 1)(4s – 5)
Correct
GUIDED PRACTICE
for Examples 1 and 2
Factor the trinomial.
3.
2h2 + 13h – 7.
SOLUTION
Because b is positive and c is negative, the factors
of c have different signs.
EXAMPLE 2
Factor when b is negative and c is positive
Factors
of 2
Factors of
–7
Possible
factorization
Middle term
when multiplied
1, 2
1, – 7
(h + 1)(2h – 7)
– 7h + 2h = 5h
1, 2
– 7, 1
(h – 7)(2h + 1)
h – 14h = – 13h
1, 2
– 1, 7
(h – 1)(2h + 7)
7h – 2h = 5h
1, 2
7, – 1
(h + 7)(2h – 1)
– h + 14h = 13h
ANSWER
2h2 + 13h – 7 = (h + 7)(2h – 1)
Correct
EXAMPLE 3
Factor when a is negative
Factor – 4x2 + 12x + 7.
SOLUTION
STEP 1
Factor – 1 from each term of the trinomial.
– 4x2 + 12x + 7 = –(4x2 – 12x – 7)
STEP 2
Factor the trinomial 4x2 – 12x – 7. Because b and c are
both negative, the factors of c must have different
signs. As in the previous examples, use a table to
organize information about the factors of a and c.
EXAMPLE 3
Factor when a is negative
Factors
of 4
Factors
of – 7
Possible
factorization
Middle term
when multiplied
1, 4
1, – 7
(x + 1)(4x – 7)
– 7x + 4x = – 3x
1, 4
7, – 1
(x + 7)(4x – 1)
– x + 28x = 27x
1, 4
– 1, 7
(x – 1)(4x + 7)
7x – 4x = 3x
1, 4
– 7, 1
(x – 7)(4x + 1)
x – 28x = – 27x
2, 2
1, – 7
(2x + 1)(2x – 7)
– 14x + 2x = – 12x
2, 2
– 1, 7
(2x – 1)(2x + 7)
14x – 2x = 12x
Correct
EXAMPLE 3
Factor when a is negative
ANSWER
– 4x2 + 12x + 7 = –(2x + 1)(2x – 7)
CHECK
You can check your factorization using a
graphing calculator. Graph
y1 = –4x2 + 12x + 7 and y2 = (2x + 1)(2x – 7).
Because the graphs coincide, you know
that your factorization is correct.
GUIDED PRACTICE
for Example 3
Factor the trinomial.
4. – 2y2 – 5y – 3
SOLUTION
STEP 1
Factor – 1 from each term of the trinomial.
– 2y2 – 5y – 3 = –(2y2 + 5y + 3)
STEP 2
Factor the trinomial 2y2 + 5y + 3. Because b and c are
both positive, the factors of c must have both
positive. Use a table to organize information about
the factors of a and c.
GUIDED PRACTICE
for Example 3
ANSWER
– 2y2 – 5y – 3 = – (y + 1)(2y + 3)
GUIDED PRACTICE
for Example 3
Factor the trinomial.
5. – 5m2 + 6m – 1
SOLUTION
STEP 1
Factor – 1 from each term of the trinomial.
– 5m2 + 6m – 1 = – (5m2 – 6m + 1)
STEP 2
Factor the trinomial 5m2 – 6m + 1. Because b is
negative and c is positive, the factors of c must be
both negative. Use a table to organize information
about the factors of a and c.
GUIDED PRACTICE
for Example 3
ANSWER
– 5m2 + 6m – 1 = – (m – 1)(5m – 1)
EXAMPLE 4
Write and solve a polynomial equation
Discus
An athlete throws a discus from an initial height of 6
feet and with an initial vertical velocity of 46 feet per
second.
a. Write an equation that gives the
height (in feet) of the discus as a
function of the time (in seconds)
since it left the athlete’s hand.
b. After how many seconds
does the discus hit the
ground?
EXAMPLE 4
Write and solve a polynomial equation
SOLUTION
a.
b.
Use the vertical motion model to write an
equation for the height h (in feet) of the discus. In
this case, v = 46 and s = 6.
h = – 16t2 + vt + s
Vertical motion model
h = – 16t2 + 46t + 6
Substitute 46 for v and 6 for s.
To find the number of seconds that pass before
the discus lands, find the value of t for which the
height of the discus is 0. Substitute 0 for h and
solve the equation for t.
EXAMPLE 4
Write and solve a polynomial equation
0 = – 16t2 + 46t + 6
Substitute 0 for h.
0 = – 2(8t2 – 23t – 3)
Factor out – 2.
0 = – 2(8t + 1)(t – 3)
Factor the trinomial. Find
factors of 8 and – 3 that produce
a middle term with a coefficient
of – 23.
8t + 1 = 0 or t – 3 = 0
t = – 1 or
8
t=3
Zero-product property
Solve for t.
EXAMPLE 4
Write and solve a polynomial equation
The solutions of the equation are – 81 and 3. A negative
solution does not make sense in this situation, so
disregard – 1 .
8
ANSWER
The discus hits the ground after 3 seconds.
Warm-Up – 9.7 and 9.8
Lesson 9.7, For use with pages 600-605
Find the product.
1. (m + 2)(m – 2)
ANSWER
m2 – 4
2. (2y – 3)2
ANSWER
4y2 – 12y + 9
Lesson 9.7, For use with pages 600-605
Find the product.
3. (s + 2t)(s – 2t)
ANSWER
s2 – 4t2
4. A football is thrown in the air at an initial height of
5 feet and an initial velocity of 16 feet per second.
After how many seconds does it hit the ground?
H(t) = - 16t2 + vt + s
ANSWER
1.25 sec
Lesson 9.8, For use with pages 606-613
1. Solve 2x2 + 11x = 21.
ANSWER
3 , –7
2
2. Factor 4x2 + 10x + 4.
ANSWER
(2x + 4)(2x + 1)
Lesson 9.8, For use with pages 606-613
3. A replacement piece of sod for a lawn has an area
of 112 square inches. The width is w and the length
is 2w – 2. What are the dimensions of the sod?
ANSWER
width: 8in., length 14 in.
Vocabulary – 9.7 and 9.8
• Perfect Square Trinomial
• (a + b)2 = a2 + 2ab + b2
• (a - b)2 = a2 - 2ab + b2
• Difference of two squares
• a2 – b2 = (a + b)(a – b)
• Factor by Grouping
• Look for this when you have 4 terms in a polynomial
• Factor out GCF from first two terms and second two
terms.
• Factor Completely
• Polynomial w/ integer coefficients that can’t be factored
any more
Notes – 9.7 and 9.8
• Before you factor a polynomial, FACTOR OUT
THE GCF IN ALL THE TERMS FIRST!
•The GCF can be a polynomial as well!!
•Try these steps to ensure a polynomial is factored
Examples 9.7 and 9.8
EXAMPLE 1
Factor the difference of two squares
Factor the polynomial.
a. y2 – 16 = y2 – 42
= (y + 4)(y – 4)
b.
25m2 – 36 = (5m)2 – 62
= (5m + 6)(5m – 6)
c. x2 – 49y2 = x2 – (7y)2
= (x + 7y)(x – 7y)
Write as a2 – b2.
Difference of two squares
pattern
Write as a2 – b2.
Difference of two squares
pattern
Write as a2 – b2.
Difference of two squares
pattern
EXAMPLE 2
Factor the difference of two squares
Factor the polynomial 8 – 18n2.
8 – 18n2 = 2(4 – 9n2)
Factor out common factor.
= 2[22 – (3n) 2]
Write 4 – 9n2 as a2 – b2.
= 2(2 + 3n)(2 – 3n)
Difference of two squares pattern
GUIDED PRACTICE
for Examples 1 and 2
Factor the polynomial.
1. 4y2 – 64 = (2y)2 – (8)2
= (2y + 8)(2y – 8)
Write as a2 – b2.
Difference of two squares
pattern
EXAMPLE 3
Factor perfect square trinomials
Factor the polynomial.
a.
n2 – 12n + 36 = n2 – 2(n 6) + 62
= (n – 6)2
b.
9x2 – 12x + 4 = (3x)2 – 2(3x 2) + 22
= (3x – 2)2
c.
4s2 + 4st + t2 = (2s)2 + 2(2s t) + t2
= (2s + t)2
Write as a2 – 2ab + b2.
Perfect square
trinomial pattern
Write as a2 – 2ab + b2.
Perfect square
trinomial pattern
Write as a2 + 2ab + b2.
Perfect square
trinomial pattern
EXAMPLE 4
Factor a perfect square trinomial
Factor the polynomial – 3y2 + 36y – 108.
– 3y2 + 36y – 108 = – 3(y2 – 12y + 36)
Factor out – 3.
= – 3(y2 – 2(y 6) + 62) Write y2 – 12y + 36
as a2 – 2ab + b2.
= – 3(y – 6)2
Perfect square
trinomial pattern
GUIDED PRACTICE
for Examples 3 and 4
Factor the polynomial.
2.
h2 + 4h + 4 = h2+2(h 2) +22
= (h + 2)2
3.
2y2 – 20y + 50 = 2(y2 – 10y +25)
Write as a2 +2ab+ b2.
Perfect square trinomial
pattern
Factor out 2
= 2[y2 –2(y 5) + 52] Write as y2 –10y+25 as
a2 –2ab+b2 .
= 2(y – 5)2
Perfect square trinomial
pattern
GUIDED PRACTICE
4.
for Examples 3 and 4
3x2 + 6xy + 3y2= 3(x2 + 2xy +y2)
Factor out 3
= 3[x2 +2(x y)+y2]
Write as x2 +2xy+ y2 as
a2 +2ab+b2 .
= 3(x + y)2
Perfect square trinomial
pattern
EXAMPLE 1
Factor out a common binomial
Factor the expression.
a.
2x(x + 4) – 3(x + 4)
b.
3y2(y – 2) + 5(2 – y)
SOLUTION
a.
b.
2x(x + 4) – 3(x + 4) = (x + 4)(2x – 3)
The binomials y – 2 and 2 – y are opposites.
Factor – 1 from 2 – y to obtain a common
binomial factor.
3y2(y – 2) + 5(2 – y) = 3y2(y – 2) – 5(y – 2) Factor – 1 from (2 – y).
= (y – 2)(3y2 – 5)
Distributive property
EXAMPLE 2
Factor by grouping
Factor the polynomial.
a
x3 + 3x2 + 5x + 15.
b.
y2 + y + yx + x
SOLUTION
a. x3 + 3x2 + 5x + 15 = (x3 + 3x2) + (5x + 15) Group terms.
= x2(x + 3) + 5(x + 3) Factor each group.
Distributive property
= (x + 3)(x2 + 5)
b.
y2 + y + yx + x = (y2 + y) + (yx + x)
= y(y + 1) + x(y + 1)
= (y + 1)(y + x)
Group terms.
Factor each group.
Distributive property
EXAMPLE 3
Factor by grouping
Factor x3 – 6 + 2x – 3x2.
SOLUTION
The terms x 3 and – 6 have no common factor. Use the
commutative property to rearrange the terms so that
you can group terms with a common factor.
x3– 6 +2x – 3x2 = x3– 3x2 +2x – 6
Rearrange terms.
= (x3 – 3x2 ) + (2x – 6)
Group terms.
= x2 (x – 3 ) + 2(x – 3)
Factor each group.
= (x – 3 ) (x2+ 2)
Distributive property
EXAMPLE 3
Factor by grouping
CHECK
Check your factorization using a
graphing calculator. Graph
y 1= x3 – 6 + 2x – 3x2 and
y2 = (x – 3)(x2 + 2) .
Because the graphs coincide,
you know that your factorization
is correct.
GUIDED PRACTICE
for Examples 1, 2 and 3
Factor the expression.
1.
x (x – 2) + (x – 2)
x (x – 2) + (x – 2) = x (x – 2) + 1(x – 2)
= (x – 2) (x + 1)
a3 + 3a2 + a + 3.
a3 + 3a2 + a + 3 = (a3 + 3a2) + (a + 3)
Factor 1 from x – 2.
Distributive property
2.
Group terms.
= a2(a + 3) + 1(a + 3)
Factor each group.
= (a2 + 1)(a + 3)
Distributive property
GUIDED PRACTICE
3.
for Examples 1, 2 and 3
y2 + 2x + yx + 2y.
SOLUTION
The terms y2 and 2x have no common factor. Use the
commutative property to rearrange the terms so that
you can group terms with a common factor.
y2 + 2x + yx + 2y = y2 + yx + 2y +2x
Rearrange terms.
= ( y2 + yx ) +( 2y +2x ) Group terms.
= y( y + x ) + 2(y +x )
Factor each group.
= (y + 2)( y + x )
Distributive property
Review – Ch. 9 – PUT
HW QUIZZES HERE
Daily Homework Quiz
For use after Lesson 9.1
If the expression is a polynomial, find its degree
and classify it by the number of
terms.Otherwise, tell why it is not a polynomial
1. m3 + n4m2 + m–2
ANSWER
2.
No; one exponent is not a whole
number.
– 3b3c4 – 4b2c+c8
ANSWER
8th degree trinomial
Daily Homework Quiz
For use after Lesson 9.1
Find the sum or difference.
3.
(3m2 –2m+9) + (m2+2m – 4)
ANSWER
4.
4m2+5
(– 4a2 + 3a – 1) – (a2 + 2a – 6)
ANSWER
–5a2 + a + 5
Daily Homework Quiz
5.
For use after Lesson 9.1
The number of dog adoptions D and cat
adoptions C can be modeled by
D = 1.35 t2 –9.8t+131 and C= 0.1t2 –3t+79 where t
represents the years since 1998. About how
many dogs and cats were adopted in 2004?
ANSWER
about 185 dogs and cats
Daily Homework Quiz
For use after Lesson 9.2
Find the product.
1. 3x(x3 – 3x2 + 2x – 4)
ANSWER
2.
3x4 –9x3+6x2 –12x
(y – 4)(2y + 5)
ANSWER
2y2 –3y –20
Daily Homework Quiz
3.
(4x + 3)(3x – 2)
ANSWER
4.
For use after Lesson 9.2
12x2+x – 6
(b2 – 2b – 1)(3b – 5)
ANSWER
3b3 – 11b2 + 7b + 5
Daily Homework Quiz
For use after Lesson 9.2
5. The dimensions of a rectangle are x+4 and
3x – 1.Write an expression to represent the
area of the rectangle.
ANSWER
3x2+11x – 4
Daily Homework Quiz
For use after Lesson 9.3
Find the product.
1. (y + 8)(y – 8)
ANSWER
2.
y2 –64
(3m – 2n)2
ANSWER
9m2 – 12 mn + 4n2
Daily Homework Quiz
3.
(2m + 5)2
ANSWER
4.
For use after Lesson 9.3
4m2 + 20m + 25
In humans, the genes for being able to roll
and not roll the tongue and R and r,
respectively. Offspring with R can roll the
tongue. If one parent is Rr and the other is
rr,what percent of the offspring will not be
able to roll the tongue?
ANSWER
50%
Daily Homework Quiz
Solve the equation.
1. (y + 5 ) (y – 9 ) = 0
ANSWER
2.
–5,9
(2n + 3 ) (n – 4 ) = 0
ANSWER
–
3 ,4
2
3. 6x2 =20x
ANSWER
0, 10
3
For use after Lesson 9.4
Daily Homework Quiz
4.
12x2 =18x
ANSWER
5.
For use after Lesson 9.4
0, 3
2
A dog jumps in the air with an initial velocity of
18 feet per second to catch a flying disc. How
long does the dog remain in the air?
ANSWER
1.125 sec
Daily Homework Quiz
Factor the trinomial.
1.
x2 – 6x –16
ANSWER
2.
y2 + 11y + 24
ANSWER
3.
(x +2) (x – 8)
(y +3) (y + 8)
x2 + x – 12
ANSWER
(x +4) (x – 3)
For use after Lesson 9.5
Daily Homework Quiz
For use after Lesson 9.5
4. Solve a2 – a = 20
ANSWER
– 4, 5
5. Each wooden slat on a set of blinds has width w
and length w + 17 . The area of one slat is 38 square
inches. What are the dimensions of a slat?
ANSWER
2 in. by 19 in
Daily Homework Quiz
For use after Lesson 9.6
Factor the trinomial.
1. – x2 + x +30
ANSWER
2.
5b2 +3b – 14
ANSWER
2.
– (x + 5) (x – 6)
(b + 2) (5b – 7)
6y2 – 13y – 5
ANSWER
(3y + 1) (2y – 5)
Daily Homework Quiz
4.
Solve 2x2 + 7x = – 3
ANSWER
5.
For use after Lesson 9.6
– 1 , –3
2
A baseball is hit into the air at an initial height of
4 feet and an initial velocity of 30 feet per second.
For how many seconds is it in the air?
ANSWER
2 sec
Daily Homework Quiz
Factor the trinomial.
1.
4m2 – n2
ANSWER
(2m – n) (2m + n)
2. x2 + 6x +9
ANSWER
3.
(x + 3)2
4y2 – 16y +16
ANSWER
(2y – 4)2
For use after Lesson 9.7
Daily Homework Quiz
4.
For use after Lesson 9.7
Solve the equation
1
+x+
=0
4
1
–
ANSWER
2
x2
5.
An apple falls from a branch 9 feet above the
ground. After how many seconds does the apple
hit the ground.
ANSWER
0.75 sec
Daily Homework Quiz
For use after Lesson 9.8
Factor the polynomial completely.
1.
40b5 – 5b3
5b3(8b2 – 1)
ANSWER
2.
x3 + 6x2 – 7x
ANSWER
3.
x(x – 1)(x + 7)
y3 + 6y2 – y – 6
ANSWER
(y + 6 )(y2 – 1)
Daily Homework Quiz
4.
Solve 2x3 + 18x2 = – 40x.
ANSWER
5.
For use after Lesson 9.8
– 5, – 4, 0
A sewing kit has a volume of 72 cubic
inches.Its dimensions are w,w + 1, and 9 – w
units.Find the dimensions of the kit.
ANSWER
8 in. by 9 in. by 1 in.