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

Examples X.X

Warm-Up – Chapter 9

Prerequisite Skills SKILL CHECK Simplify the expression. 7. 3x +(– 6x) –3 x ANSWER 8. 5 + 4x + 2 4x + 7 ANSWER 9. 4(2x – 1) + x 9x – 4 ANSWER – 7x – 4 ANSWER 10. – (x + 4) – 6 x

Prerequisite Skills SKILL CHECK Simplify the expression. 11. (3xy)3 27x3y3 ANSWER 12. xy2 xy3 x2y5 ANSWER x15 ANSWER 13. (x5)3 –x3 ANSWER 14. (– x)3

Vocabulary – 9.1 Degree of a polynomial Monomial Term with highest degree Leading Coefficient Coefficient of the highest degree term Binomial Polynomial with 2 terms Trinomial Polynomial with 3 terms 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

Notes – 9.1 - Polynomials CLASSIFYING POLYNOMIALS What is NOT a polynomial? Terms with: Negative exponents Fractional exponents 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 Combine like terms Remember to include the signs of the coefficients! SUBTRACTING POLYNOMIALS Use distributive property first!!

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.

Identify and classify polynomials 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. 5th degree binomial Yes 7bc3 + 4b4c No; variable exponent n– 2 – 3 6n4 – 8n 2nd degree trinomial 2x2 + x – 5 0 degree monomial 9 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)

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 GUIDED PRACTICE Rewrite a polynomial for Examples 1,2, and 3 Write 5y – 2y2 + 9 so that the exponents decrease from left to right. Identify the degree and leading coefficient of the polynomial. 1. 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 GUIDED PRACTICE Identify and classify polynomials for Examples 1,2, and 3 for Example 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.

EXAMPLE 3 GUIDED PRACTICE Add polynomials for Examples 1,2, and 3 for Example Find the sum. 3. a. (2x3 + 4x – x) + (4x2 +3x3 – 6)

EXAMPLE 3 GUIDED PRACTICE Add polynomials 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) = (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) 4n2 + 5 – (– 2n2 + 2n – 4) 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 GUIDED PRACTICE Subtract polynomials for Examples 4 and 5 Find the difference. 4. a. (4x2 + 7x) – ( 5x2 + 4x – 9) (4x2 – 7x ) – (5x2 – 4x – 9) = 4x2 – 7x – 5x2 + 4x + 9 = (4x2 – 5x2) + (– 7x – 4x) + 9 = –x2 – 11x + 9

Solve a multi-step problem EXAMPLE 5 Solve a multi-step problem BASEBALL ATTENDNCE Look back at Example 5. Find The difference in attendance at National and American League baseball games in 2001. 5. 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 x3 + 7x2 + 8x + 5 2. Simplify r2s rs3. ANSWER r3s4 3. Simplify 2x(3x + 2) ANSWER 6x2 + 4x

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 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

Multiply a monomial and a polynomial 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. 3x2 x – 4 3x 2 – 8 – 12x 2x 3x2 x – 4 3x 2 ANSWER The product is 3x2 + 2x – 12x – 8, or 3x2 – 10x – 8.

GUIDED PRACTICE for Examples 1 and 2 Find the product. x(7x2 +4) 1 SOLUTION x(7x2 +4) Write product. = x(7x2 )+x(4) Distributive property = 7x3+4x 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. 3 6a a 2a2 2a 1 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)

GUIDED PRACTICE for Examples 1 and 2 STEP 2 Make a table of products. – 5 –n 20n 4n2 4n –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

Multiply binomials using the FOIL pattern 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.

GUIDED PRACTICE for Examples 3, 4, and 5 Find the product. (x2 + 2x +1)(x + 2) 4 SOLUTION x3 + 4x2 + 5x + 2

GUIDED PRACTICE for Examples 3, 4, and 5 Find the product. 5 (3y2 –y + 5)(2y – 3) 5 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.

= (4b)(b) + (4b)(– 2) + (–5)(b) + (–5)(– 2) GUIDED PRACTICE for Examples 3, 4, and 5 Find the product. (4b –5)(b – 2) 6 SOLUTION = (4b)(b) + (4b)(– 2) + (–5)(b) + (–5)(– 2) Write products of terms. = 4b2 – 8b – 5b + 10 Multiply. = 4b2 – 13b + 10 Combine like terms.

Standardized Test Practice EXAMPLE 6 Standardized Test Practice The dimensions of a rectangle are x + 3 and x + 2. Which expression represents the area of the rectangle? x2 + 6 A x2 + 5x + 6 B x2 + 6x + 6 C x2 + 6x D 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.

Standardized Test Practice EXAMPLE 6 Standardized Test Practice = x2 + 5x + 6 Combine like terms. ANSWER The correct answer is B. A D C 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.

Warm-Up – 9.3

Lesson 9.3, For use with pages 569-574 Find the product. 1. (x + 7)(x + 7) 3. (x + 7)(x - 7) ANSWER x2 + 14x + 49 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

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

Use the square of a binomial pattern EXAMPLE 1 Find the product. a. (3x + 4)2 =(3x)2 + 2(3x)(4) + 42 Square of a binomial pattern = 9x2 + 24x + 16 Simplify. b. (5x – 2y)2 = (5x)2 – 2(5x)(2y) + (2y)2 Square of a binomial pattern =25x2 – 20xy + 4y2 Simplify.

GUIDED PRACTICE for Example 1 Find the product. 1. (x + 3)2 = (x)2 + 2(x)(3) + (3)2 Square of a binomial pattern = x2 + 6x + 9 Simplify. 2. (2x + 1)2 = (2x)2 + 2(2x)(1) + (1)2 Square of a binomial pattern =4x2 + 4x + 1 Simplify.

GUIDED PRACTICE for Example 1 3. (4x – y)2 = (4x)2 – 2(4x)(y) + (y)2 Square of a binomial pattern = 16x2 – 8xy + y2 Simplify. 4. (3m +n)2 = (3m)2 + 2(3m)(n) + (n)2 Square of a binomial pattern = 9m2 + 6mn + n2 Simplify.

Use the sum and difference pattern EXAMPLE 2 Use the sum and difference pattern Find the product. a. (t + 5)(t – 5) = t2 – 52 Sum and difference pattern = t2 – 25 Simplify. b. (3x + y)(3x – y) = (3x)2 – y2 Sum and difference pattern = 9x2 – y2 Simplify.

GUIDED PRACTICE for Example 2 Find the product. 5. (x + 10)(x – 10) Sum and difference pattern = x2 – 100 Simplify. 6. (2x + 1)(2x – 1) = (2x)2 – 12 Sum and difference pattern = 4x2 – 1 Simplify. 7. (x + 3y)(x – 3y) = (x)2 – (3y)2 Sum and difference pattern = x2 – 9y2 Simplify.

Use special products and mental math 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 Sum and difference pattern = 900 – 16 Evaluate powers. = 884 Simplify.

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.

8. Describe how you can use special product to find 212. GUIDED PRACTICE for Examples 3 and 4 8. 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 Square of a binomial pattern = 421 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 Find the binomials that multiply to get x2 - 9 ANSWER (x + 3)(x - 3) 2. (2a – 5b)(2a + 5b) ANSWER 4a2 – 25b2

Lesson 9.4, For use with pages 575-580 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. ANSWER about 10,070 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 Height of the object Above the ground t = time (in seconds) v = initial velocity (in feet/second) s = initial height (in feet)

Examples 9.4

Use the zero-product property EXAMPLE 1 Use the zero-product property Solve (x – 4)(x + 2) = 0. (x – 4)(x + 2) = 0 Write original equation. x – 4 = 0 or x + 2 = 0 Zero-product property x = 4 or x = – 2 Solve for x. ANSWER The solutions of the equation are 4 and –2.

1. Solve the equation (x – 5)(x – 1) = 0. GUIDED PRACTICE for Example 1 1. Solve the equation (x – 5)(x – 1) = 0. (x – 5)(x – 1) = 0 Write original equation. x – 5 = 0 or x – 1 = 0 Zero-product property x = 5 or x = 1 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 for Example 2 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)

Solve an equation by factoring EXAMPLE 3 Solve an equation by factoring Solve 2x2 + 8x = 0. 2x2 + 8x = 0. Write original equation. 2x(x + 4) = 0 Factor left side. 2x = 0 or x + 4 = 0 Zero-product property x = 0 or x = – 4 Solve for x. ANSWER The solutions of the equation are 0 and – 4.

Solve an equation by factoring EXAMPLE 4 Solve an equation by factoring Solve 6n2 = 15n. 6n2 – 15n = 0 Subtract 15n from each side. 3n(2n – 5) = 0 Factor left side. 3n = 0 or 2n – 5 = 0 Zero-product property n = 5 2 n = 0 or Solve for n. ANSWER The solutions of the equation are 0 and 5 2 .

The solutions of the equation are 0 and – 5. GUIDED PRACTICE for Examples 3 and 4 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 Zero-product property a = 0 or a = – 5 Solve for x. ANSWER The solutions of the equation are 0 and – 5.

The solutions of the equation are 0 and 3. GUIDED PRACTICE for Examples 3 and 4 4. 3s2 – 9s = 0. 3s2 – 9s = 0 Write original equation. 3s(s – 3) = 0 Factor left side. 3s = 0 or s – 3 = 0 Zero-product property s= 0 or s = 3 Solve for x. ANSWER The solutions of the equation are 0 and 3.

The solutions of the equation are 0 and 1 2 . GUIDED PRACTICE for Examples 3 and 4 5. 4x2 = 2x. 4x2 = 2x Write original equation. 4x2 – 2x = 0 Subtract 2x from each side. 2x(2x – 1) = 0 Factor left side. 2x = 0 or 2x – 1 = 0 Zero-product property x = 1 2 x = 0 or Solve for x. ANSWER The solutions of the equation are 0 and 1 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?

Solve a multi-step problem 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.

Solve a multi-step problem 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 –8t + 7 = 0 Zero-product property t = 0 or t = 0.875 Solve for t. ANSWER The armadillo lands on the ground 0.875 second after the armadillo jumps.

Write a model for the armadillo’s height above the ground. 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 4t – 3 = 0 Zero-product property t = 0 or t = 0.75 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 y2 – 8y + 15 2. (y + 3)( y + 5) 4. (2y + 3)( y + 5) ANSWER y2 + 8y + 15 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

Factor when b and c are positive 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 9, 2 6, 3 18 + 1 = 19 9 + 2 = 11 6 + 3 = 9 Correct sum

Factor when b and c are positive 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.

Find two positive factors of 2 whose sum is 3. Make an organized list. 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. Correct sum Factors of 2 Sum of factors 1, 2 1 + 2 = 3

EXAMPLE 1 GUIDED PRACTICE Factor when b and c are positive for Example 1 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)

Find two positive factors of 10 whose sum is 7. 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. Correct sum Factors of 10 Sum of factors 10, 1 2, 5 10 + 1 = 11 2 + 5 = 7

EXAMPLE 1 GUIDED PRACTICE Factor when b and c are positive for Example 1 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)

Find two positive factors of 14 whose sum is 9. 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. Correct sum Factors of 14 Sum of factors 14, 1 7, 2 14 + 1 = 15 7 + 2 = 9

EXAMPLE 1 GUIDED PRACTICE Factor when b and c are positive for Example 1 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)

Factor when b is negative and c is positive 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 –4,–2 –8 + (–1) = –9 –4 + (–2) = –6 Correct sum ANSWER n2 – 6n + 8 = (n – 4)( n – 2)

Factor when b is positive and c is negative 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 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)

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. Correct sum Factors of 2 Sum of factors –3, –1 –3 + (–1) = –4 ANSWER x2 – 4x + 3 = (x – 3)( x – 1)

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 –4, –3 –12 + (–1) = –13 –6 + (–2) = –8 –4 + (–3) = –7 –6, –2 Correct sum ANSWER t2 – 8t + 12 = (t – 6)( t – 2)

Because c is negative, p and q must have different signs. 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. –5, 4 –5 + 4 = –1 5, –4 5 – 4 = 1 Factors of 20 Sum of factors –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)

Because c is negative, p and q must have different signs. 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. –4, 4 –4 + 4 = 0 Factors of 16 Sum of factors –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)

Solve a polynomial equation EXAMPLE 4 Solve a polynomial equation Solve the equation x2 + 3x = 18. x2 + 3x = 18 Write original equation. x2 + 3x – 18 = 0 Subtract 18 from each side. (x + 6)(x – 3) = 0 Factor left side. x + 6 = 0 or x – 3 = 0 Zero-product property x = – 6 or x = 3 Solve for x. ANSWER The solutions of the equation are – 6 and 3.

Solve a polynomial equation for Example 4 GUIDED PRACTICE Solve a polynomial equation for Example 4 8. Solve the equation s2 – 2s = 24. s2 – 2s = 24. Write original equation. s2 – 2s – 24 = 0 Subtract 24 from each side. (s + 4)(s – 6) = 0 Factor left side. s + 4 = 0 or s – 6 = 0 Zero product property s = – 4 or s = 6 Solve for x. ANSWER 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.

Solve a polynomial equation 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. ANSWER The banner cannot have a negative width, so the width is 3 feet.

The banner cannot have a negative width, so the width is 2 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 Formula for area of a rectangle 20 = (4 + w + 4) w Substitute 20 for A and (4 + w + 4) for l. 20 = w2 + 8w Simplify 0 = w2 + 8w – 20 Subtract 20 from each side. 0 = (w + 10)(w – 2) Factor right side. w + 10 = 0 or w – 2 = 0 Zero-product property w = – 10 or w = 2 Solve for w. ANSWER 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 Factor and solve the polynomials. Find the product. 1. (3c + 3)(2c – 3) 3. x2 – x – 6 = 0 ANSWER 6c2 – 3c – 9 ANSWER (x-3)(x+2) x = 3 or x= -2 2. (2y + 3)(2y + 1) 4. x2 + 13x = -36 ANSWER 4y2 + 8y + 3 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.

Factor when b is negative and c is positive EXAMPLE 1 Factor when b is negative and c is positive Factor 2x2 – 7x + 3. – x – 6x = – 7x (x – 3)(2x – 1) -3, -1 1, 2 – 3x – 2x = – 5x (x – 1)(2x – 3) – 1, – 3 1,2 Middle term when multiplied Possible factorization Factors of 3 Factors of 2 Correct 2x2 – 7x + 3 =(x – 3)(2x – 1) ANSWER

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.

Factor when b is negative and c is positive EXAMPLE 2 Factor when b is negative and c is positive n – 15n = –14n (n – 5)(3n + 1) – 5, 1 1, 3 – n + 15n = 14n (n + 5)(3n – 1) 5, – 1 5n – 3n = 2n (n – 1)(3n + 5) –1, 5 – 5n + 3n = – 2n (n + 1)(3n – 5) 1, –5 Middle term when multiplied Possible factorization Factors of –5 Factors of 3 Correct 3n2 + 14n – 5 = (n + 5)(3n – 1) ANSWER

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 ANSWER 2t + 6t = 8t (t + 2)(3t + 2) 2, 2 1, 3 t + 12t = 13t (t + 4)(3t + 1) 4, 1 4t + 3t = 7t (t + 1)(3t + 4) 1, 4 Middle term when multiplied Possible factorization Factors of 4 Factors of 3 Correct 3t2 + 8t + 4 = (t + 2)(3t + 2) ANSWER

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 for Examples 1 and 2 – 10s – 2s = – 12s – 1, – 5 2, 2 – s – 20s = – 21s (s – 5)(4s – 1) – 5, – 1 1, 4 – 5s – 4s = – 9s (s – 1)(4s – 5) Middle term when multiplied Possible factorization Factors of 5 Factors of 4 Correct 4s2 – 9s + 5 = (s – 1)(4s – 5) ANSWER

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.

Factor when b is negative and c is positive EXAMPLE 2 Factor when b is negative and c is positive – h + 14h = 13h (h + 7)(2h – 1) 7, – 1 1, 2 7h – 2h = 5h (h – 1)(2h + 7) – 1, 7 h – 14h = – 13h (h – 7)(2h + 1) – 7, 1 – 7h + 2h = 5h (h + 1)(2h – 7) 1, – 7 Middle term when multiplied Possible factorization Factors of – 7 Factors of 2 Correct 2h2 + 13h – 7 = (h + 7)(2h – 1) ANSWER

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.

Factor when a is negative EXAMPLE 3 Factor when a is negative 14x – 2x = 12x (2x – 1)(2x + 7) – 1, 7 2, 2 – 14x + 2x = – 12x (2x + 1)(2x – 7) 1, – 7 x – 28x = – 27x (x – 7)(4x + 1) – 7, 1 1, 4 7x – 4x = 3x (x – 1)(4x + 7) – x + 28x = 27x (x + 7)(4x – 1) 7, – 1 – 7x + 4x = – 3x (x + 1)(4x – 7) Middle term when multiplied Possible factorization Factors of – 7 of 4 Correct

EXAMPLE 3 Factor when a is negative 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

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. 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.

Write and solve a polynomial equation EXAMPLE 4 Write and solve a polynomial equation 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.

Write and solve a polynomial equation 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 Zero-product property t = – 1 8 or t = 3 Solve for t.

EXAMPLE 4 Write and solve a polynomial equation The solutions of the equation are – 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 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. ANSWER 1.25 sec

Lesson 9.8, For use with pages 606-613 1. Solve 2x2 + 11x = 21. ANSWER 3 2 , –7 2. Factor 4x2 + 10x + 4. ANSWER (2x + 4)(2x + 1)

Lesson 9.8, For use with pages 606-613 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. 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

Factor the difference of two squares EXAMPLE 1 Factor the difference of two squares Factor the polynomial. a. y2 – 16 = y2 – 42 Write as a2 – b2. = (y + 4)(y – 4) Difference of two squares pattern b. 25m2 – 36 = (5m)2 – 62 Write as a2 – b2. = (5m + 6)(5m – 6) Difference of two squares pattern c. x2 – 49y2 = x2 – (7y)2 Write as a2 – b2. = (x + 7y)(x – 7y) Difference of two squares pattern

Factor the difference of two squares 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 Write as a2 – b2. = (2y + 8)(2y – 8) Difference of two squares pattern

Factor perfect square trinomials EXAMPLE 3 Factor the polynomial. 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 = (3x)2 – 2(3x 2) + 22 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. = (2s + t)2 Perfect square trinomial pattern

Factor a perfect square trinomial 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 Write as a2 +2ab+ b2. = (h + 2)2 Perfect square trinomial pattern 3. 2y2 – 20y + 50 = 2(y2 – 10y +25) 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 for Examples 3 and 4 4. 3x2 + 6xy + 3y2 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

Factor out a common binomial EXAMPLE 1 Factor out a common binomial Factor the expression. 2x(x + 4) – 3(x + 4) a. 3y2(y – 2) + 5(2 – y) b. SOLUTION 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) Factor – 1 from (2 – y). = (y – 2)(3y2 – 5) Distributive property

y2 + y + yx + x = (y2 + y) + (yx + x) b. = y(y + 1) + x(y + 1) EXAMPLE 2 Factor by grouping Factor the polynomial. x3 + 3x2 + 5x + 15. a y2 + y + yx + x b. SOLUTION x3 + 3x2 + 5x + 15 = (x3 + 3x2) + (5x + 15) a. Group terms. = x2(x + 3) + 5(x + 3) Factor each group. = (x + 3)(x2 + 5) Distributive property y2 + y + yx + x = (y2 + y) + (yx + x) b. Group terms. = y(y + 1) + x(y + 1) Factor each group. = (y + 1)(y + x) Distributive property

EXAMPLE 3 Factor by grouping 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– 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

Check your factorization using a graphing calculator. Graph y and y EXAMPLE 3 Factor by grouping 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) = x (x – 2) + 1(x – 2) Factor 1 from x – 2. = (x – 2) (x + 1) Distributive property 2. a3 + 3a2 + a + 3. a3 + 3a2 + a + 3 = (a3 + 3a2) + (a + 3) Group terms. = a2(a + 3) + 1(a + 3) Factor each group. = (a2 + 1)(a + 3) Distributive property

GUIDED PRACTICE for Examples 1, 2 and 3 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 No; one exponent is not a whole number. ANSWER 2. – 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) 4m2+5 ANSWER 4. (– 4a2 + 3a – 1) – (a2 + 2a – 6) ANSWER –5a2 + a + 5

Daily Homework Quiz For use after Lesson 9.1 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 cats ANSWER

Daily Homework Quiz For use after Lesson 9.2 Find the product. 1. 3x(x3 – 3x2 + 2x – 4) 3x4 –9x3+6x2 –12x ANSWER 2. (y – 4)(2y + 5) ANSWER 2y2 –3y –20

Daily Homework Quiz For use after Lesson 9.2 3. (4x + 3)(3x – 2) ANSWER 12x2+x – 6 4. (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) y2 –64 ANSWER 2. (3m – 2n)2 ANSWER 9m2 – 12 mn + 4n2

Daily Homework Quiz For use after Lesson 9.3 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%

Daily Homework Quiz For use after Lesson 9.4 Solve the equation. 1. (y + 5 ) (y – 9 ) = 0 ANSWER – 5 , 9 2. (2n + 3 ) (n – 4 ) = 0 ANSWER 3 2 – , 4 3. 6x2 =20x ANSWER 10 3 0,

Daily Homework Quiz For use after Lesson 9.4 4. 12x2 =18x ANSWER 3 2 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

Daily Homework Quiz For use after Lesson 9.5 Factor the trinomial. 1. x2 – 6x –16 ANSWER (x +2) (x – 8) 2. y2 + 11y + 24 ANSWER (y +3) (y + 8) 3. x2 + x – 12 ANSWER (x +4) (x – 3)

Daily Homework Quiz For use after Lesson 9.5 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

Daily Homework Quiz For use after Lesson 9.6 Factor the trinomial. 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)

Daily Homework Quiz For use after Lesson 9.6 4. Solve 2x2 + 7x = – 3 ANSWER – 1 2 , –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

Daily Homework Quiz For use after Lesson 9.7 Factor the trinomial. 1. 4m2 – n2 ANSWER (2m – n) (2m + n) 2. x2 + 6x +9 ANSWER (x + 3)2 3. 4y2 – 16y +16 ANSWER (2y – 4)2

Daily Homework Quiz For use after Lesson 9.7 4. Solve the equation x2 + x + = 0 1 4 ANSWER – 1 2 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 ANSWER 5b3(8b2 – 1) 2. x3 + 6x2 – 7x ANSWER x(x – 1)(x + 7) 3. y3 + 6y2 – y – 6 ANSWER (y + 6 )(y2 – 1)

Daily Homework Quiz For use after Lesson 9.8 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