Download - Unit 6 – Chapter 9
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Unit 6 – Chapter 9
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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
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Warm-Up – X.X
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Vocabulary – X.X• Holder
• Holder 2
• Holder 3
• Holder 4
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Notes – X.X – LESSON TITLE.• Holder•Holder•Holder•Holder•Holder
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Examples X.X
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Warm-Up – Chapter 9
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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
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Prerequisite Skills SKILL CHECK
Simplify the expression.
13. (x5)3
14. (– x)3
11. (3xy)3 27x3y3ANSWER
12. xy2 xy3 x2y5ANSWER
x15ANSWER
–x3ANSWER
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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
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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)
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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!
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Examples 9.1
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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
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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.
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EXAMPLE 3 Add polynomials
Find the sum.
a. (2x3 – 5x2 + x) + (2x2 + x3 – 1)
b. (3x2 + x – 6) + (x2 + 4x + 10)
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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) =
= 4x2 + 5x + 4
(3x2 + x2) + (x + 4x) + (– 6 + 10)
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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
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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
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EXAMPLE 3 Add polynomials
a. (2x3 + 4x – x) + (4x2 +3x3 – 6)
Find the sum.3.
GUIDED PRACTICE for Example for Examples 1,2, and 3
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EXAMPLE 3 Add polynomials
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
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EXAMPLE 4 Subtract polynomials
Find the difference.
a. (4n2 + 5) – (– 2n2 + 2n – 4)
b. (4x2 – 3x + 5) – (3x2 – x – 8)
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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 – 3x2) + (– 3x + x) + (5 + 8)
= x2 – 2x + 13
4x2 – 3x + 5 – 3x2 + x + 8
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EXAMPLE 4 Subtract polynomials
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
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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.
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EXAMPLE 5 Solve a multi-step problem
ANSWER
About 7,320,000 people attended Major League Baseball games in 2001.
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Warm-Up – 9.2
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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
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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
Lesson 9.2, For use with pages 561-568
ANSWER x2 + 3x +2
4. Simplify (x + 1)(x + 2)
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Vocabulary – 9.2• Polynomial
• Monomial or Sum of monomials with multiple terms
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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
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Examples 9.2
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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
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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
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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.
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GUIDED PRACTICE for Examples 1 and 2
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
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GUIDED PRACTICE for Examples 1 and 2
Find the product.2 (a +3)(2a +1)
ANSWER
The product is 2a2 + a + 6a + 3, or 2a2 + 7a + 3.
Make a table of products.
36a
a2a2a
3
2a 1
SOLUTION
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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)
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GUIDED PRACTICE for Examples 1 and 2
ANSWER
The product is 4n2 + 20n – n – 5, or 4n2 + 19n – 5.
– 5–n
20n4n24n
–1
n 5
STEP 2
Make a table of products.
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EXAMPLE 3 Multiply polynomials vertically
Find the product (b2 + 6b – 7)(3b – 4).
SOLUTION 3b3 + 14b2 – 45b + 28
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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
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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.
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GUIDED PRACTICE for Examples 3, 4, and 5
Find the product.
SOLUTION
(x2 + 2x +1)(x + 2)4
x3 + 4x2 + 5x + 2
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GUIDED PRACTICE for Examples 3, 4, and 5
Find the product.
Write product.
= 3y2(2y – 3) – y(2y – 3) + 5(2y – 3)
= 6y3 – 9y2 – 2y2 + 3y + 10y – 15
Distributive property
Distributive property
= 6y3 – 11y2 + 13y – 15 Combine like terms.
(3y2 –y + 5)(2y – 3)
SOLUTION
(3y2 –y + 5)(2y – 3)5
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GUIDED PRACTICE for Examples 3, 4, and 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.
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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
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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.
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Warm-Up – 9.3
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Lesson 9.3, For use with pages 569-574
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
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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?
Lesson 9.3, For use with pages 569-574
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Vocabulary – 9.3• Binomial
• Polynomial with 2 terms
• Trinomial
• Polynomial with 3 terms
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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
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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
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Examples 9.4
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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
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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
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GUIDED PRACTICE for Example 1
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
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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
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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) 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
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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)
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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.
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Solve a multi-step problemEXAMPLE 4
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.
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Solve a multi-step problemEXAMPLE 4
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.
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Solve a multi-step problemEXAMPLE 4
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
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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.
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GUIDED PRACTICE for Examples 3 and 4
(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.
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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
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.
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GUIDED PRACTICE for Examples 3 and 4
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
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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.
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Warm-Up – 9.4
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Lesson 9.4, For use with pages 575-580
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)
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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.
Lesson 9.4, For use with pages 575-580
1. Write down an equation with values for X and Y that make the following equation true:
X * Y = 0
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Vocabulary – 9.4• Roots
• Solutions for x when a function = 0
• Also where the function crosses the X axis
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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.
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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)
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Examples 9.4
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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
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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
x = 5
Zero-product property
Solve for x.
ANSWER
The solutions of the equation are 5 and 1.
or
or x – 1 = 0
x = 1
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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)
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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
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GUIDED PRACTICE for Example 2
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)
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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.
Zero-product property
Factor left side.
Write original equation.
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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.
Zero-product property
Factor left side.
Subtract 15n from each side.
ANSWER
The solutions of the equation are 0 and52 .
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GUIDED PRACTICE for Examples 3 and 4
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.
Zero-product property
Factor left side.
Write original equation.
3. a2 + 5a = 0.
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GUIDED PRACTICE for Examples 3 and 4
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.
Zero-product property
Factor left side.
Write original equation.
4. 3s2 – 9s = 0.
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GUIDED PRACTICE for Examples 3 and 4
5. 4x2 = 2x.
4x2 – 2x = 0
2x(2x – 1) = 0
2x = 0
x = 0
2x – 1 = 0
x =12
or
or Solve for x.
Zero-product property
Factor left side.
Subtract 2x from each side.
ANSWER
The solutions of the equation are 0 and12 .
4x2 = 2x Write original equation.
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ARMADILLO
EXAMPLE 5 Solve a multi-step problem
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?
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SOLUTION
EXAMPLE 5 Solve a multi-step problem
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.
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EXAMPLE 5 Solve a multi-step problem
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.
Zero-product property
Factor right side.
Substitute 0 for h.
ANSWER
The armadillo lands on the ground 0.875 second after the armadillo jumps.
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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 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.
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GUIDED PRACTICE for Example 5
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.
Zero-product property
Factor right side.
Substitute 0 for h.
ANSWER
The armadillo lands on the ground 0.75 second after the armadillo jumps.
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Warm-Up – 9.5
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Lesson 9.5, For use with pages 582-589
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
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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
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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.
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Examples 9.5
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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
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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
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SOLUTION
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.
Correct sum
Factors of 2 Sum of factors
1, 2 1 + 2 = 3
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Factor when b and c are positiveEXAMPLE 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)
GUIDED PRACTICE for Example 1
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SOLUTION
Find two positive factors of 10 whose sum is 7. Make an organized list.
GUIDED PRACTICE for Example 1
Factor the trinomial
2. a2 + 7a + 10.
Correct sum
Factors of 10 Sum of factors
10, 1
2, 5
10 + 1 = 11
2 + 5 = 7
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Factor when b and c are positiveEXAMPLE 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)
GUIDED PRACTICE for Example 1
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SOLUTION
Find two positive factors of 14 whose sum is 9. Make an organized list.
GUIDED PRACTICE for Example 1
Factor the trinomial
3. t2 + 9t + 14.
Correct sum
Factors of 14 Sum of factors
14, 1
7, 2
14 + 1 = 15
7 + 2 = 9
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Factor when b and c are positiveEXAMPLE 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)
GUIDED PRACTICE for Example 1
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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)
Factors of 8 Sum of factors
–8,–1
–4,–2
–8 + (–1) = –9
–4 + (–2) = –6 Correct sum
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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)
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4. x2 – 4x + 3.
Because b is negative and c is positive, p and q must both be negative.
ANSWER
x2 – 4x + 3 = (x – 3)( x – 1)
GUIDED PRACTICE for Examples 2 and 3
Factor the trinomial
Correct sum
Factors of 2 Sum of factors
–3, –1 –3 + (–1) = –4
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5. t2 – 8t + 12.
Because b is negative and c is positive, p and q must both be negative.
GUIDED PRACTICE for Examples 2 and 3
Factor the trinomial
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)
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6. m2 + m – 20.
Because c is negative, p and q must have different signs.
GUIDED PRACTICE for Examples 2 and 3
Factor the trinomial
–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)
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7. w2 + 6w – 16.
Because c is negative, p and q must have different signs.
GUIDED PRACTICE for Examples 2 and 3
Factor the trinomial
–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)
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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.
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EXAMPLE 4 Solve a polynomial equation
8. Solve the equation s2 – 2s = 24.
Write original equation.
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.
GUIDED PRACTICE for Example 4
s2 – 2s = 24.
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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.
STEP 1Draw a diagram of two banners together.
SOLUTION
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EXAMPLE 5 Solve a polynomial equation
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
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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)
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.
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.
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Warm-Up – 9.6
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Lesson 9.6, For use with pages 592-599
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
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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
Lesson 9.6, For use with pages 592-599
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Vocabulary – 9.6• Trinomial
• Polynomial with 3 terms
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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
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Examples 9.6
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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.
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EXAMPLE 1 Factor when b is negative and c is positive
– 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.
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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.
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EXAMPLE 2 Factor when b is negative and c is positive
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
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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.
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GUIDED PRACTICE for Examples 1 and 2
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
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GUIDED PRACTICE for Examples 1 and 2
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.
Factor the trinomial.
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GUIDED PRACTICE for Examples 1 and 2
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
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GUIDED PRACTICE for Examples 1 and 2
3. 2h2 + 13h – 7.
SOLUTION
Because b is positive and c is negative, the factors of c have different signs.
Factor the trinomial.
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EXAMPLE 2 Factor when b is negative and c is positive
– 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
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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.
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EXAMPLE 3 Factor when a is negative
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
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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
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GUIDED PRACTICE for Example 3
Factor the trinomial.
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.
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GUIDED PRACTICE for Example 3
ANSWER
– 2y2 – 5y – 3 = – (y + 1)(2y + 3)
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GUIDED PRACTICE for Example 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.
Factor the trinomial.
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GUIDED PRACTICE for Example 3
ANSWER
– 5m2 + 6m – 1 = – (m – 1)(5m – 1)
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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.
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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.
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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 Zero-product property
t = – 18
Solve for t.
or t – 3 = 0
or t = 3
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The solutions of the equation are – and 3. A negative solution does not make sense in this situation, so disregard – .
18
18
EXAMPLE 4 Write and solve a polynomial equation
ANSWER
The discus hits the ground after 3 seconds.
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Warm-Up – 9.7 and 9.8
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Lesson 9.7, For use with pages 600-605
Find the product.
1. (m + 2)(m – 2)
2. (2y – 3)2
ANSWER m2 – 4
ANSWER 4y2 – 12y + 9
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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
Lesson 9.7, For use with pages 600-605
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Lesson 9.8, For use with pages 606-613
1. Solve 2x2 + 11x = 21.
2. Factor 4x2 + 10x + 4.
ANSWER (2x + 4)(2x + 1)
ANSWER32
, –7
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ANSWER width: 8in., length 14 in.
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.
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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
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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
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Examples 9.7 and 9.8
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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
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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.
Difference of two squares pattern
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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
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EXAMPLE 3 Factor perfect square trinomials
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 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
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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)
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GUIDED PRACTICE for Examples 3 and 4
Factor the polynomial.
= (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)
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GUIDED PRACTICE for Examples 3 and 4
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)
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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).
Distributive property
Factor the expression.
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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.
Distributive property
Factor the polynomial.
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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
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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
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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.
Distributive property
a3 + 3a2 + a + 3 =
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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.
Distributive property
y2 + yx + 2y +2x
= ( y2 + yx ) +( 2y +2x )
= y( y + x ) + 2(y +x ) Factor each group.
= (y + 2)( y + x )
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Review – Ch. 9 – PUT HW QUIZZES HERE
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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
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Daily Homework Quiz For use after Lesson 9.1
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
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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 catsANSWER
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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
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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
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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
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Daily Homework Quiz For use after Lesson 9.3
Find the product.
1. (y + 8)(y – 8)
y2 –64ANSWER
2. (3m – 2n)2
ANSWER 9m2 – 12 mn + 4n2
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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%
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Daily Homework Quiz For use after Lesson 9.4
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.
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Daily Homework Quiz For use after Lesson 9.4
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
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Daily Homework Quiz For use after Lesson 9.5
Factor the trinomial.
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)
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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
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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)
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Daily Homework Quiz For use after Lesson 9.6
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
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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
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Daily Homework Quiz For use after Lesson 9.7
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
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Daily Homework Quiz For use after Lesson 9.8
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)
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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