unit 6 – chapter 9. unit 6 chapter 8 review and chap. 8 skills section 9.1 – adding and...

Unit 6 – Chapter 9

Unit 6• 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.• Holder•Holder•Holder•Holder•Holder

Examples X.X

Warm-Up – Chapter 9

7. 3x +(– 6x)

Prerequisite Skills SKILL CHECK

Simplify the expression.

8. 5 + 4x + 2

–3 xANSWER

4x + 7ANSWER

9. 4(2x – 1) + x 9x – 4ANSWER

10. – (x + 4) – 6 x – 7x – 4ANSWER

Prerequisite Skills SKILL CHECK

Simplify the expression.

13. (x5)3

14. (– x)3

11. (3xy)3 27x3y3ANSWER

12. xy2 xy3 x2y5ANSWER

x15ANSWER

–x3ANSWER

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


Notes – 9.1 - Polynomials• CLASSIFYING POLYNOMIALS •What is NOT a polynomial? Terms with:

1. Negative exponents2. Fractional exponents3. 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 terms2. Remember to include the signs of the

coefficients!•SUBTRACTING POLYNOMIALS

1. Use distributive property first!!2. Combine like terms3. Remember to include the signs of the

coefficients!

Examples 9.1

Rewrite a polynomial EXAMPLE 1

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.

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.

15x – x3 + 3

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.

EXAMPLE 2 Identify and classify polynomials

5th degree binomialYes7bc3 + 4b4c

No; variable exponentn– 2 – 3

No; variable exponent6n4 – 8n

2nd degree trinomialYes2x2 + x – 5

0 degree monomialYes9

Classify by degree and number of terms

Is it a polynomial?Expression

a.

b.c.d.e.

EXAMPLE 3 Add polynomials

Find the sum.

a. (2x3 – 5x2 + x) + (2x2 + x3 – 1)

b. (3x2 + x – 6) + (x2 + 4x + 10)


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) =

= 4x2 + 5x + 4

(3x2 + x2) + (x + 4x) + (– 6 + 10)

Rewrite a polynomial EXAMPLE 1

SOLUTION

Consider the degree of each of the polynomial’s terms.

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

5y –2y2 + 9

Write 5y – 2y2 + 9 so that the exponents decrease from left to right. Identify the degree and leading coefficient of the polynomial.

1.

GUIDED PRACTICE for Examples 1,2, and 3

Degree is 1Degree is 2

Degree is 0

EXAMPLE 2 Identify and classify polynomials

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.

2.

SOLUTION

y3 – 4y + 3 is a polynomial. 3 degree trinomial.

GUIDED PRACTICE for Examplefor Examples 1,2, and 3


a. (2x3 + 4x – x) + (4x2 +3x3 – 6)

Find the sum.3.

GUIDED PRACTICE for Example for Examples 1,2, and 3


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) =

= 8x3 + 4x2 + 2x – 6

(5x3 + 3x3) + (4x2) + (4x –2x) + (– 6)

GUIDED PRACTICE for Example for Examples 1,2, and 3

EXAMPLE 4 Subtract polynomials

Find the difference.

a. (4n2 + 5) – (– 2n2 + 2n – 4)

b. (4x2 – 3x + 5) – (3x2 – x – 8)


SOLUTION

a. (4n2 + 5) 4n2 + 5

– (– 2n2 + 2n – 4) 2n2 – 2n + 4

6n2 – 2n + 9

b. (4x2 – 3x + 5) – (3x2 – x – 8) =

= (4x2 – 3x2) + (– 3x + x) + (5 + 8)

= x2 – 2x + 13

4x2 – 3x + 5 – 3x2 + x + 8


a. (4x2 + 7x) – ( 5x2 + 4x – 9)

GUIDED PRACTICE for Examples 4 and 5

Find the difference.4.

(4x2 – 7x ) – (5x2 – 4x – 9) =

= (4x2 – 5x2) + (– 7x – 4x) + 9

= –x2 – 11x + 9

4x2 – 7x – 5x2 + 4x + 9

EXAMPLE 5 Solve a multi-step problem

BASEBALL ATTENDNCE Look back at Example 5. FindThe difference in attendance at National and AmericanLeague baseball games in 2001.

5.

M = (– 488t2 + 5430t + 24,700) – (– 318t2 + 3040t + 25,600)

= (– 488t2 + 318t2) + (5430t – 3040t) + (24,700 – 25,600)

= – 170t2 + 2390t – 900

– 488t2 + 5430t + 24,700 + 318t2 – 3040t – 25,600=

7320= Substitute 6 for t in the model, because 2001 is 6 years after 1995.


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).

ANSWER –18a + 2b

ANSWER r3s4

2. Simplify r2s rs3.

ANSWER 6x2 + 4x

3. Simplify 2x(3x + 2)

ANSWER x3 + 7x2 + 8x + 5

4. Simplify x2(x+1) + 2x(3x+3) + 2x +5

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


ANSWER x2 + 3x +2

4. Simplify (x + 1)(x + 2)

Vocabulary – 9.2• Polynomial

• Monomial or Sum of monomials with multiple terms

Notes – 9.2 – Multiply Polynomials• Multiplying Polynomials is like using the distributive 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

Multiply polynomials using a tableEXAMPLE 2

Find the product (x – 4)(3x + 2).

STEP 1

Write subtraction as addition in each polynomial.

(x – 4)(3x + 2) = [x + (– 4)](3x + 2)

SOLUTION

Multiply polynomials using a tableEXAMPLE 2

ANSWER

The product is 3x2 + 2x – 12x – 8, or 3x2 – 10x – 8.

3x2x

– 4

3x 2

– 8– 12x

2x3x2x

– 4

3x 2

STEP 2

Make a table of products.


Find the product.1 x(7x2 +4)

x(7x2 +4) Write product.

= x(7x2 )+x(4) Distributive property

= 7x3+4x Product of powers property

SOLUTION


Find the product.2 (a +3)(2a +1)

ANSWER

The product is 2a2 + a + 6a + 3, or 2a2 + 7a + 3.


36a

a2a2a

3

2a 1

SOLUTION


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)


ANSWER

The product is 4n2 + 20n – n – 5, or 4n2 + 19n – 5.

– 5–n

20n4n24n

–1

n 5

STEP 2


EXAMPLE 3 Multiply polynomials vertically

Find the product (b2 + 6b – 7)(3b – 4).

SOLUTION 3b3 + 14b2 – 45b + 28

Multiply polynomials horizontallyEXAMPLE 4

Find the product (2x2 + 5x – 1)(4x – 3).

FOIL PATTERN The letters of the word FOIL can helpyou to remember how to use the distributive property tomultiply binomials. The letters should remind you of thewords First, Outer, Inner, and Last.

(2x + 3)(4x + 1)

First Outer Inner Last

= 8x2 + 2x + 12x + 3

Solution: Multiply everything and get= (2x2)(4x) + (2x2)(-3) + (5x)(4x) + (5x)(-3) + (-1)(4x) + (-1)(-3)= 8x3 + 14x2 – 19x + 3

Multiply binomials using the FOIL pattern

EXAMPLE 5

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.

GUIDED PRACTICE for Examples 3, 4, and 5

Find the product.

SOLUTION

(x2 + 2x +1)(x + 2)4

x3 + 4x2 + 5x + 2


Find the product.

Write product.

= 3y2(2y – 3) – y(2y – 3) + 5(2y – 3)

= 6y3 – 9y2 – 2y2 + 3y + 10y – 15

Distributive property


= 6y3 – 11y2 + 13y – 15 Combine like terms.

(3y2 –y + 5)(2y – 3)

SOLUTION

(3y2 –y + 5)(2y – 3)5


Find the product.

SOLUTION

(4b –5)(b – 2)6

= (4b)(b) + (4b)(– 2) + (–5)(b) + (–5)(– 2) Write products of terms.

= 4b2 – 8b – 5b + 10 Multiply.

= 4b2 – 13b + 10 Combine like terms.

SOLUTION

EXAMPLE 6 Standardized Test Practice

The dimensions of a rectangle are x + 3 and x + 2. Which expression represents the area of the rectangle?

Area = length width Formula for area of a rectangle

= (x + 3)(x + 2) Substitute for length and width.

= x2 + 2x + 3x + 6 Multiply binomials.

x2 + 6 A x2 + 6xDx2 + 5x + 6B x2 + 6x + 6C

EXAMPLE 6

= x2 + 5x + 6 Combine like terms.

ANSWER

The correct answer is B. A DCB

Standardized Test Practice

CHECKYou 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.

Warm-Up – 9.3


Find the product.

1. (x + 7)(x + 7)

2. (x - 7)(x - 7)

ANSWER x2 + 14x + 49

ANSWER x2 - 14x + 49

3. (x + 7)(x - 7)

ANSWER x2 – 49

2. (3x – 1)(3x + 2)

ANSWER 9x2 + 3x – 2

ANSWER 15x2 + 46x + 16; 1344 m2

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?


Vocabulary – 9.3• Binomial


• Trinomial


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 - b2

•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 Square of a binomial pattern

= 9x2 + 24x + 16 Simplify.

b. (5x – 2y)2 Square of a binomial pattern

=25x2 – 20xy + 4y2 Simplify.

=(3x)2 + 2(3x)(4) + 42

= (5x)2 – 2(5x)(2y) + (2y)2

GUIDED PRACTICE for Example 1

Find the product.

1. (x + 3)2Square of a binomial pattern

= x2 + 6x + 9 Simplify.

2. (2x + 1)2 Square of a binomial pattern

=4x2 + 4x + 1 Simplify.

= (x)2 + 2(x)(3) + (3)2

= (2x)2 + 2(2x)(1) + (1)2


3. (4x – y)2Square of a binomial pattern

= 16x2 – 8xy + y2 Simplify.

4. (3m +n)2 Square of a binomial pattern

= 9m2 + 6mn + n2 Simplify.

= (4x)2 – 2(4x)(y) + (y)2

= (3m)2 + 2(3m)(n) + (n)2

Use the sum and difference pattern EXAMPLE 2

Find the product.

a. (t + 5)(t – 5) Sum and difference pattern

= t2 – 25 Simplify.

b. (3x + y)(3x – y) Sum and difference pattern

= 9x2 – y2 Simplify.

= t2 – 52

= (3x)2 – y2


Find the product.

5. (x + 10)(x – 10) Sum and difference pattern

= x2 – 100 Simplify.

6. (2x + 1)(2x – 1) Sum and difference pattern

= 4x2 – 1 Simplify.

7. (x + 3y)(x – 3y) Sum and difference pattern

= x2 – 9y2 Simplify.

= x2 – 102

= (2x)2 – 12

= (x)2 – (3y)2

Use special products and mental mathEXAMPLE 3

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 Write as product of difference and sum.

= 302 – 42 Sum and difference pattern

= 900 – 16 Evaluate powers.

= 884 Simplify.

= (30 – 4)(30 + 4)

Solve a multi-step problemEXAMPLE 4

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.


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.


SOLUTION

Notice that the Punnett square shows 4 possible gene combinations of the offspring. Of these combinations, 3 result in black patches.

STEP 1

ANSWER

75% of the possible gene combinations result in black patches.


STEP 2Model 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

Solve a multi-step problem EXAMPLE 4

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.


(20 + 1)2 = (20)2 + 2(20)(1) + 12 Square of a binomial pattern

= 421 Simplify.

Use the square of binomial pattern to find the product (20 +1)2.

8. Describe how you can use special product to find 212.


BORDER COLLIES

Look back at Example 4. What percent of the possible gene combinations of the offspring result in red patches?

SOLUTION

Notice that the Punnett square shows 4 possible gene combinations of the offspring. Of these combinations, 1 result in red patches.

STEP 1

ANSWER

25% of the possible gene combinations result in redpatches.


STEP 2Model 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


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


1. Find the GCF of 12 and 28.

ANSWER 4

2. (2a – 5b)(2a + 5b)

ANSWER 4a2 – 25b2

3. Find the binomials that multiply to get x2 - 9

ANSWER (x + 3)(x - 3)

ANSWER about 10,070

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?

3.


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

t = time (in seconds)Height of the objectAbove the ground

v = initial velocity (in feet/second)

s = initial height (in feet)

Examples 9.4

Use the zero-product property EXAMPLE 1

Solve (x – 4)(x + 2) = 0.

(x – 4)(x + 2) = 0 Write original equation.

x – 4 = 0 x = 4

Zero-product property

Solve for x.

ANSWER

The solutions of the equation are 4 and –2.

oror x + 2 = 0

x = – 2


1. Solve the equation (x – 5)(x – 1) = 0.

(x – 5)(x – 1) = 0 Write original equation.

x – 5 = 0

x = 5


Solve for x.

ANSWER

The solutions of the equation are 5 and 1.

or

or x – 1 = 0

x = 1

SOLUTION

EXAMPLE 2 Find the greatest common monomial factor

Factor out the greatest common monomial factor.

a. 12x + 42y

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

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)

SOLUTION

Factor out the greatest common monomial factor.

b. 4x4 + 24x3


2. Factor out the greatest common monomial factorfrom 14m + 35n.

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.

SOLUTION

ANSWER

14m + 35n = 7(2m + 5n)

EXAMPLE 3 Solve an equation by factoring

Solve 2x2 + 8x = 0.

2x2 + 8x = 0.

2x(x + 4) = 0

2x = 0

x = 0

or x + 4 = 0

or x = – 4

ANSWER

The solutions of the equation are 0 and – 4.

Solve for x.


Factor left side.

Write original equation.

EXAMPLE 4 Solve an equation by factoring

Solve 6n2 = 15n.

6n2 – 15n = 0

3n(2n – 5) = 0

3n = 0

n = 0

2n – 5 = 0

n =52

or

or Solve for n.


Factor left side.

Subtract 15n from each side.

ANSWER

The solutions of the equation are 0 and52 .


Solve the equation.

a2 + 5a = 0

a(a + 5) = 0

a = 0

a = 0

or a + 5 = 0

or a = – 5

ANSWER

The solutions of the equation are 0 and – 5.

Solve for x.


Factor left side.


3. a2 + 5a = 0.


3s2 – 9s = 0

3s(s – 3) = 0

3s = 0

s= 0

or s – 3 = 0

or s = 3

ANSWER

The solutions of the equation are 0 and 3.

Solve for x.


Factor left side.


4. 3s2 – 9s = 0.


5. 4x2 = 2x.

4x2 – 2x = 0

2x(2x – 1) = 0

2x = 0

x = 0

2x – 1 = 0

x =12

or

or Solve for x.


Factor left side.

Subtract 2x from each side.

ANSWER

The solutions of the equation are 0 and12 .

4x2 = 2x Write original equation.

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?

SOLUTION


STEP 1

Write a model for the armadillo’s height above the ground.

h = – 16t2 + vt + s

h = – 16t2 + 14t + 0

h = – 16t2 + 14t

Vertical motion model

Substitute 14 for v and 0 for s.

Simplify.


STEP 2Substitute 0 for h. When the armadillo lands, its height above the ground is 0 feet. Solve for t.

0 = – 16t2 + 14t

0 = 2t(–8t + 7)

2t = 0

t = 0

–8t + 7 = 0

t = 0.875

or

or Solve for t.


Factor right side.

Substitute 0 for h.

ANSWER

The armadillo lands on the ground 0.875 second after the armadillo jumps.


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 1Write a model for the armadillo’s height above the ground.

h = – 16t2 + vt + s

h = – 16t2 + 12t + 0

Vertical motion model

Substitute 12 for v and 0 for s.

h = – 16t2 + 12t Simplify.


STEP 2Substitute 0 for h. When the armadillo lands, its height above the ground is 0 feet. Solve for t.

0 = – 16t2 + 12t

0 = – 4t(4t – 3)

– 4t = 0

t = 0

4t – 3 = 0

t = 0.75

or

or Solve for t.


Factor right side.

Substitute 0 for h.

ANSWER

The armadillo lands on the ground 0.75 second after the armadillo jumps.

Warm-Up – 9.5


Find the product.

1. (y + 3)(y - 5)

2. (y + 3)( y + 5)

ANSWER y2 - 2x – 15

ANSWER y2 + 8y + 15

3. (y - 3)( y - 5)

ANSWER 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) Polynomialx2 + 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

Factor when b and c are positiveEXAMPLE 1

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

9, 2

6, 3

18 + 1 = 19

9 + 2 = 11

6 + 3 = 9

Correct sum

Factor when b and c are positive EXAMPLE 1

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)

(x + 9)(x + 2) Multiply binomials.= x2 + 2x + 9x + 18

Simplify.

CHECK

= x2 + 11x + 18

SOLUTION



Factor the trinomial

1. x2 + 3x + 2.

Correct sum


1, 2 1 + 2 = 3



ANSWER

x2 + 3x + 2 = (x + 2)(x + 1)


SOLUTION




2. a2 + 7a + 10.

Correct sum


10, 1

2, 5

10 + 1 = 11

2 + 5 = 7



ANSWER

a2 + 7a + 10 = (a + 5)(a + 2)


SOLUTION




3. t2 + 9t + 14.

Correct sum


14, 1

7, 2

14 + 1 = 15

7 + 2 = 9



ANSWER

t2 + 9t + 14 = (t + 7)(t + 2)


Factor when b is negative and c is positiveEXAMPLE 2

Factor n2 – 6n + 8.

Because b is negative and c is positive, p and q must both be negative.

ANSWER

n2 – 6n + 8 = (n – 4)( n – 2)


–8,–1

–4,–2

–8 + (–1) = –9

–4 + (–2) = –6 Correct sum

Factor when b is positive and c is negativeEXAMPLE 3

Factor y2 + 2y – 15.

Because c is negative, p and q must have different signs.

Factors of –15 Sum of factors

–15, 1

–5, 3

–15 + 1 = –14

15 + (–1) = 14

–5 + 3 = –2

15, –1

5, –3 5 + (–3) = 2 Correct sum

ANSWER y2 + 2y – 15 = (y + 5)( y – 3)

4. x2 – 4x + 3.


ANSWER

x2 – 4x + 3 = (x – 3)( x – 1)



Correct sum


–3, –1 –3 + (–1) = –4

5. t2 – 8t + 12.





–12, –1

–4, –3

–12 + (–1) = –13

–6 + (–2) = –8

–4 + (–3) = –7

–6, –2 Correct sum

ANSWER t2 – 8t + 12 = (t – 6)( t – 2)

6. m2 + m – 20.




–5, 4 –5 + 4 = –1

5, –4 5 – 4 = 1


–20, 1

–10, 2

–20 + 1 = –19

–1 + 20 = 19

– 10 + 2 = –8

–1, 20

–2,10 – 2 +10 = 8

Correct sum

ANSWER m2 + m – 20 = (m + 5)( m – 4)

7. w2 + 6w – 16.




–4, 4 –4 + 4 = 0


–16, 1

–8, 2

–16 + 1 = –15

16 + (–1) = 15

– 8 + 2 = –6

16, –1

8, –2 8 + (–2) = 6 Correct sum

ANSWER w2 + 6w – 16 = (w + 8)( w – 2)

EXAMPLE 4 Solve a polynomial equation

Solve the equation x2 + 3x = 18.

Write original equation.x2 + 3x = 18

Subtract 18 from each side.x2 + 3x – 18 = 0

Factor left side.(x + 6)(x – 3) = 0

Zero-product propertyx – 3 = 0x + 6 = 0

Solve for x.x = 3

or

orx = – 6

ANSWER The solutions of the equation are – 6 and 3.


8. Solve the equation s2 – 2s = 24.


Subtract 24 from each side.s2 – 2s – 24 = 0

Factor left side.(s + 4)(s – 6) = 0

Zero product propertys – 6 = 0s + 4 = 0

Solve for x.s = 6

or

ors = – 4

ANSWER The solutions of the equation are – 4 and 6.


s2 – 2s = 24.


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.

STEP 1Draw a diagram of two banners together.

SOLUTION


STEP 2Write an equation using the fact that the area of 2 banners is 2(16.5) = 33 square feet. Solve the equation for w.

Formula for area of a rectangleA = l w

Substitute 33 for A and (4 + w + 4) for l.

Simplify and subtract 33 from each side.0 = w2 + 8w – 33

Factor right side.0 = (w + 11)(w – 3)

Zero-product propertyw + 11 = 0 or w – 3 = 0

33 = (4 + w + 4) w

Solve for w.w = – 11

ANSWERThe banner cannot have a negative width, so the width is 3 feet.

w = 3or

Write an equation using the fact that the area of 2 banners is 2(10) = 20 square feet. Solve the equation for w.

Formula for area of a rectangleA = l wSubstitute 20 for A and (4 + w + 4) for l.

Simplify20 = w2 + 8w Subtract 20 from each side.

0 = (w + 10)(w – 2)


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.

20 = (4 + w + 4) w

0 = w2 + 8w – 20Factor right side.

Solve for w.w = – 10Zero-product propertyw + 10 = 0 or w – 2 = 0

or w = 2

ANSWERThe banner cannot have a negative width, so the width is 2 feet.

Warm-Up – 9.6


Find the product.

1. (3c + 3)(2c – 3)

2. (2y + 3)(2y + 1)

ANSWER 6c2 – 3c – 9

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

ANSWER 0.75 sec

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


Vocabulary – 9.6• Trinomial


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 workFactors 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.


– x – 6x = – 7x(x – 3)(2x – 1)-3, -11, 2

– 3x – 2x = – 5x(x – 1)(2x – 3)– 1, – 31,2

Middle term

when multiplied

Possible

factorization

Factors

of 3

Factors of 2

Correct

2x2 – 7x + 3 =(x – 3)(2x – 1)ANSWER

Factor 2x2 – 7x + 3.

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.


n – 15n = –14n(n – 5)(3n + 1)– 5, 11, 3

– n + 15n = 14n(n + 5)(3n – 1)5, – 11, 3

5n – 3n = 2n(n – 1)(3n + 5)–1, 51, 3

– 5n + 3n = – 2n(n + 1)(3n – 5)1, –51, 3

Middle term

when multiplied

Possible

factorization

Factors of –5

Factors of 3

Correct

3n2 + 14n – 5 = (n + 5)(3n – 1)ANSWER


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.


3t2 + 8t + 4 = (t + 2)(3t + 2)

ANSWER

2t + 6t = 8t(t + 2)(3t + 2)2, 21, 3

t + 12t = 13t(t + 4)(3t + 1)4, 11, 3

4t + 3t = 7t(t + 1)(3t + 4)1, 41, 3

Middle term

when multiplied

Possible

factorization

Factors of 4

Factors of 3

Correct


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.



4s2 – 9s + 5 = (s – 1)(4s – 5)

ANSWER

– 10s – 2s = – 12s(2s – 1)(2s – 5)– 1, – 52, 2

– s – 20s = – 21s(s – 5)(4s – 1)– 5, – 11, 4

– 5s – 4s = – 9s(s – 1)(4s – 5)– 1, – 51, 4

Middle term

when multiplied

Possible

factorization

Factors of 5

Factors of 4

Correct


3. 2h2 + 13h – 7.

SOLUTION

Because b is positive and c is negative, the factors of c have different signs.



– h + 14h = 13h(h + 7)(2h – 1)7, – 11, 2

7h – 2h = 5h(h – 1)(2h + 7)– 1, 71, 2

h – 14h = – 13h(h – 7)(2h + 1)– 7, 11, 2

– 7h + 2h = 5h(h + 1)(2h – 7)1, – 71, 2

Middle term

when multiplied

Possible

factorization

Factors of – 7

Factors of 2

Correct

2h2 + 13h – 7 = (h + 7)(2h – 1)

ANSWER

SOLUTION

EXAMPLE 3 Factor when a is negative

Factor – 4x2 + 12x + 7.

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.


14x – 2x = 12x(2x – 1)(2x + 7)– 1, 72, 2

– 14x + 2x = – 12x(2x + 1)(2x – 7)1, – 72, 2

x – 28x = – 27x(x – 7)(4x + 1)– 7, 11, 4

7x – 4x = 3x(x – 1)(4x + 7)– 1, 71, 4

– x + 28x = 27x(x + 7)(4x – 1)7, – 11, 4

– 7x + 4x = – 3x(x + 1)(4x – 7)1, – 71, 4

Middle term

when multiplied

Possible

factorization

Factors

of – 7

Factors

of 4

Correct


ANSWER

– 4x2 + 12x + 7 = –(2x + 1)(2x – 7)

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.

CHECK



4. – 2y2 – 5y – 3

SOLUTION

STEP 1Factor – 1 from each term of the trinomial.

– 2y2 – 5y – 3 = –(2y2 + 5y + 3)STEP 2Factor 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.


ANSWER

– 2y2 – 5y – 3 = – (y + 1)(2y + 3)


5. – 5m2 + 6m – 1

SOLUTION

STEP 1Factor – 1 from each term of the trinomial.

– 5m2 + 6m – 1 = – (5m2 – 6m + 1)

STEP 2Factor 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.



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.

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.

a.

After how many seconds does the discus hit the ground?

b.


SOLUTION

a. 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.

b. 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.


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 Zero-product property

t = – 18

Solve for t.

or t – 3 = 0

or t = 3

The solutions of the equation are – and 3. A negative solution does not make sense in this situation, so disregard – .

18

18


ANSWER

The discus hits the ground after 3 seconds.

Warm-Up – 9.7 and 9.8


Find the product.

1. (m + 2)(m – 2)

2. (2y – 3)2

ANSWER m2 – 4

ANSWER 4y2 – 12y + 9

ANSWER 1.25 sec

Find the product.

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

4.

3. (s + 2t)(s – 2t)

ANSWER s2 – 4t2



1. Solve 2x2 + 11x = 21.

2. Factor 4x2 + 10x + 4.

ANSWER (2x + 4)(2x + 1)

ANSWER32

, –7

ANSWER width: 8in., length 14 in.


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?

3.

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.


Write as a2 – b2.


EXAMPLE 2 Factor the difference of two squares

Factor the polynomial 8 – 18n2.

8 – 18n2 = 2(4 – 9n2)

= 2[22 – (3n) 2]

= 2(2 + 3n)(2 – 3n)

Factor out common factor.

Write 4 – 9n2 as a2 – b2.




1. 4y2 – 64 = (2y)2 – (8)2

= (2y + 8)(2y – 8)

Write as a2 – b2.


EXAMPLE 3 Factor perfect square trinomials


a.

n2 – 12n + 36 = n2 – 2(n 6) + 62 Write as a2 – 2ab + b2.

= (n – 6)2 Perfect square trinomial pattern

b. 9x2 – 12x + 4 Write as a2 – 2ab + b2.

= (3x – 2)2 Perfect square trinomial pattern

c. 4s2 + 4st + t2 = (2s)2 + 2(2s t) + t2 Write as a2 + 2ab + b2.

= (3x)2 – 2(3x 2) + 22

= (2s + t)2 Perfect square trinomial pattern

EXAMPLE 4 Factor a perfect square trinomial

Factor the polynomial – 3y2 + 36y – 108.

– 3y2 + 36y – 108 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

= – 3(y2 – 12y + 36)



= (h + 2)2 Perfect square trinomial pattern

Write as a2 +2ab+ b2.2. h2 + 4h + 4

3. 2y2 – 20y + 50

Write as y2 –10y+25 as a2 –2ab+b2 .

Factor out 2

= 2(y – 5)2 Perfect square trinomial pattern

= 2[y2 –2(y 5) + 52]

= h2+2(h 2) +22

= 2(y2 – 10y +25)


4. 3x2 + 6xy + 3y2

Write as x2 +2xy+ y2 as a2 +2ab+b2 .

Factor out 3

= 3(x + y)2 Perfect square trinomial pattern

= 3[x2 +2(x y)+y2]

= 3(x2 + 2xy +y2)

Factor out a common binomialEXAMPLE 1

2x(x + 4) – 3(x + 4)a.

SOLUTION

3y2(y – 2) + 5(2 – y)b.

2x(x + 4) – 3(x + 4) = (x + 4)(2x – 3)a.

The binomials y – 2 and 2 – y are opposites. Factor – 1 from 2 – y to obtain a common binomial factor.

b.

3y2(y – 2) + 5(2 – y) = 3y2(y – 2) – 5(y – 2)

= (y – 2)(3y2 – 5)

Factor – 1 from (2 – y).


Factor the expression.

Factor by groupingEXAMPLE 2

x3 + 3x2 + 5x + 15.a y2 + y + yx + xb.

SOLUTION

= x2(x + 3) + 5(x + 3)= (x + 3)(x2 + 5)

x3 + 3x2 + 5x + 15 = (x3 + 3x2) + (5x + 15)a.

y2 + y + yx + x = (y2 + y) + (yx + x)b.= y(y + 1) + x(y + 1)= (y + 1)(y + x)

Group terms.

Factor each group.Distributive property

Group terms.

Factor each group.



Factor by groupingEXAMPLE 3

Factor 6 + 2x .x3 – 3x2–

SOLUTION

The terms x and – 6 have no common factor. Use the commutative property to rearrange the terms so that you can group terms with a common factor.

3

x3– 3x2 +2x – 6 x3– 6 +2x – 3x2 = Rearrange terms.

(x3 – 3x2 ) + (2x – 6)= Group terms.

x2 (x – 3 ) + 2(x – 3) = Factor each group.

(x – 3 ) (x2+ 2) = Distributive property

Factor by grouping EXAMPLE 3

CHECK

Check your factorization using a graphing calculator. Graph

y and

y

Because the graphs coincide, you know that your factorization is correct.

1

= (x – 3)(x2 + 2) .2

6 + 2x = x3 – – 3x2

GUIDED PRACTICE for Examples 1, 2 and 3

Factor the expression.

1. x (x – 2) + (x – 2)

x (x – 2) + (x – 2) = Factor 1 from x – 2.x (x – 2) + 1(x – 2)

= (x – 2) (x + 1) Distributive property

2. a3 + 3a2 + a + 3.

(a3 + 3a2) + (a + 3)

= a2(a + 3) + 1(a + 3)

= (a2 + 1)(a + 3)

Group terms.

Factor each group.


a3 + 3a2 + a + 3 =

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.

GUIDED PRACTICE for Examples 1, 2 and 3

3. y2 + 2x + yx + 2y.

SOLUTION

y2 + 2x + yx + 2y = Rearrange terms.

Group terms.


y2 + yx + 2y +2x

= ( y2 + yx ) +( 2y +2x )

= y( y + x ) + 2(y +x ) Factor each group.

= (y + 2)( y + x )

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

No; one exponent is not a whole number.

ANSWER

2. – 3b3c4 – 4b2c+c8

ANSWER 8th degree trinomial


Find the sum or difference.

3. (3m2 –2m+9) + (m2+2m – 4)

4m2+5ANSWER

4. (– 4a2 + 3a – 1) – (a2 + 2a – 6)

ANSWER –5a2 + a + 5


5. 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?

about 185 dogs and catsANSWER


Find the product.

1. 3x(x3 – 3x2 + 2x – 4)

3x4 –9x3+6x2 –12x ANSWER

2. (y – 4)(2y + 5)

ANSWER 2y2 –3y –20


3. (4x + 3)(3x – 2)

ANSWER 12x2+x – 6

4. (b2 – 2b – 1)(3b – 5)

ANSWER 3b3 – 11b2 + 7b + 5


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


Find the product.

1. (y + 8)(y – 8)

y2 –64ANSWER

2. (3m – 2n)2

ANSWER 9m2 – 12 mn + 4n2


3. (2m + 5)2

ANSWER 4m2 + 20m + 25

4. 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%


1. (y + 5 ) (y – 9 ) = 0

ANSWER – 5 , 9

2. (2n + 3 ) (n – 4 ) = 0

ANSWER 32

– , 4

3. 6x2 =20x

ANSWER103

0,

Solve the equation.


4. 12x2 =18x

ANSWER32

0,

5. 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



1. x2 – 6x –16

2. y2 + 11y + 24

ANSWER (x +2) (x – 8)

3. x2 + x – 12

ANSWER (y +3) (y + 8)

ANSWER (x +4) (x – 3)


4. Solve a2 – a = 20

ANSWER – 4, 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?

5.

ANSWER 2 in. by 19 in



1. – x2 + x +30

ANSWER – (x + 5) (x – 6)

2. 5b2 +3b – 14

ANSWER (b + 2) (5b – 7)

2. 6y2 – 13y – 5

ANSWER (3y + 1) (2y – 5)


4. Solve 2x2 + 7x = – 3

ANSWER – 12

, –3

5. 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



1. 4m2 – n2

ANSWER (2m – n) (2m + n)

2. x2 + 6x +9

ANSWER (x + 3)2

3. 4y2 – 16y +16

ANSWER (2y – 4)2


4. Solve the equation

x2 + x + = 014

ANSWER –12

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


1. 40b5 – 5b3

Factor the polynomial completely.

ANSWER 5b3(8b2 – 1)

2. x3 + 6x2 – 7x

ANSWER x(x – 1)(x + 7)

3. y3 + 6y2 – y – 6

ANSWER (y + 6 )(y2 – 1)


4. Solve 2x3 + 18x2 = – 40x.

ANSWER – 5, – 4, 0

5. 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.

8 in. by 9 in. by 1 in.ANSWER

unit 6 – chapter 9. unit 6 chapter 8 review and chap. 8 skills section 9.1 – adding and...

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