5 Minute Warm-Up

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8-1 Factors and Greatest Common Factors

5 Minute Warm-UpDirections: Simplify each problem. Write

theanswer in standard form.1. (4x + 7)(– 2x)2. -4x2(3x2 + 2x – 6)3. (x + 6)(x + 9)4. (-4y + 5)(-7 – 3y)5. (-8x3 + x – 9x2 + 2) + (8x2 – 2x + 4)6. (6x2 – x + 3) – (-2x + x2 – 7)

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Warm Up

1. 50, 6 2. 105, 7

3. List the factors of 28.

Tell whether each number is prime or composite. If the number is composite, write it as the product of two numbers.

no yes

±1, ±2, ±4, ±7,

Tell whether the second number is a factor of the first number

±14, ±28

4. 11 5. 98 composite; 49 2prime

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Write the prime factorization of numbers.

Find the GCF of monomials.

Objectives

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The whole numbers that are multiplied to find a product are called factors of that product. A number is divisible by its factors.

You can use the factors of a number to write the number as a product. The number 12 can be factored several ways.

Factorizations of 12

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Factorizations of 12

The order of factors does not change the product, but there is only one example below that cannot be factored further. The circled factorization is the prime factorization because all the factors are prime numbers. The prime factors can be written in any order, and except for changes in the order, there is only one way to write the prime factorization of a number.

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A prime number has exactly two factors, itself and 1. The number 1 is not prime because it only has one factor.

Remember!

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Example 1: Writing Prime Factorizations

Write the prime factorization of 98.

Method 1 Factor tree Method 2 Ladder diagramChoose any two factorsof 98 to begin. Keep finding factors until each branch ends in a prime factor.

Choose a prime factor of 98 to begin. Keep dividing by prime factors until the quotient is 1.

98 = 2 7 7

982 49

7 7

9849

71

2

77

98 = 2 7 7

The prime factorization of 98 is 2 7 7 or 2 72.

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Check It Out! Example 1

Write the prime factorization of each number.

a. 40

402 20

2 10

2 5

333

11

b. 33

40 = 23 5 33 = 3 11

The prime factorization of 40 is 2 2 2 5 or 23 5.

The prime factorization of 33 is 3 11.

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Factors that are shared by two or more whole numbers are called common factors. The greatest of these common factors is called the greatest common factor, or GCF.

Factors of 12: 1, 2, 3, 4, 6, 12

Factors of 32: 1, 2, 4, 8, 16, 32

Common factors: 1, 2, 4

The greatest of the common factors is 4.

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Example 2A: Finding the GCF of Numbers

Find the GCF of each pair of numbers.

100 and 60

factors of 100: 1, 2, 4, 5, 10, 20, 25, 50, 100

factors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

The GCF of 100 and 60 is 20.

List all the factors.

Circle the GCF.

Method 1 List the factors.

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Example 2B: Finding the GCF of Numbers

Find the GCF of each pair of numbers.

26 and 52

26 = 2 13

52 = 2 2 13


Align the common factors.

2 13 = 26


Method 2 Prime factorization.

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You can also find the GCF of monomials that include variables. To find the GCF of monomials, write the prime factorization of each coefficient and write all powers of variables as products. Then find the product of the common factors.

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Example 3A: Finding the GCF of Monomials

Find the GCF of each pair of monomials.

15x3 and 9x2

15x3 = 3 5 x x x

9x2 = 3 3 x x

3 x x = 3x2

Write the prime factorization of each coefficient and write powers as products.


Find the product of the common factors.

The GCF of 3x3 and 6x2 is 3x2.

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Example 3B: Finding the GCF of Monomials


8x2 and 7y3

The GCF 8x2 and 7y is 1.

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If two terms contain the same variable raised to different powers, the GCF will contain that variable raised to the lower power.

Helpful Hint

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Check It Out! Example 3a


18g2 and 27g3

18g2 = 2 3 3 g g

27g3 = 3 3 3 g g g

3 3 g g

The GCF of 18g2 and 27g3 is 9g2.



Find the product of the common factors.

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Check It Out! Example 3b


16a6 and 9b

9b = 3 3 b

16a6 = 2 2 2 2 a a a a a a



There are no common factors other than 1.

The GCF of 16a6 and 7b is 1.

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Adrianne is shopping for a CD storage unit. She has 36 CDs by pop music artists and 48 CDs by country music artists. She wants to put the same number of CDs on each shelf without putting pop music and country music CDs on the same shelf. If Adrianne puts the greatest possible number of CDs on each shelf, how many shelves does her storage unit need?

The 36 pop and 48 country CDs must be divided into groups of equal size. The number of CDs in each row must be a common factor of 36 and 48.

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Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36

Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

Find the common factors of 36 and 48.


Check It Out! Example 4 Continued

The greatest possible number of CDs on each shelf is 12. Find the number of shelves of each type of CDs when Adrianne puts the greatest number of CDs on each shelf.

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36 pop CDs12 CDs per shelf = 3 shelves

48 country CDs 12 CDs per shelf

= 4 shelves

When the greatest possible number of CD types are on each shelf, there are 7 shelves in total.

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8-2 Factoring by GCF5 Minute Warm-Up


1. 50 2. 84

Find the GCF of each pair monomial.

3. 18 and 75 4. 20 and 36

5. 12x and 28x3 6. 27x2 and 45x3y2

7. Cindi is planting a rectangular flower bed with 40 orange flower and 28 yellow flowers. She wants to plant them so that each row will have the same number of plants but of only one color. How many rows will Cindi need if she puts the greatest possible number of plants in each row?

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8-2 Factoring by GCF

Factor polynomials by using the greatest common factor.

Objective

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Recall that the Distributive Property states that ab + ac =a(b + c). The Distributive Property allows you to “factor” out the GCF of the terms in a polynomial to write a factored form of the polynomial.

A polynomial is in its factored form when it is written as a product of monomials and polynomials that cannot be factored further. The polynomial 2(3x – 4x) is not fully factored because the terms in the parentheses have a common factor of x.

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Example 1A: Factoring by Using the GCFFactor each polynomial. Check your answer.

2x2 – 4

2x2 = 2 x x 4 = 2 2

2

Find the GCF.

The GCF of 2x2 and 4 is 2.Write terms as products using the

GCF as a factor.2x2 – (2 2)

2(x2 – 2)

Check 2(x2 – 2)

2x2 – 4

Multiply to check your answer.The product is the original

polynomial.

Use the Distributive Property to factor out the GCF.

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Aligning common factors can help you find the greatest common factor of two or more terms.

Writing Math

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Example 1B: Factoring by Using the GCFFactor each polynomial. Check your answer.

8x3 – 4x2 – 16x

2x2(4x) – x(4x) – 4(4x)

4x(2x2 – x – 4)

4x(2x2 – x – 4)

8x3 – 4x2 – 16x

8x3 = 2 2 2 x x x4x2 = 2 2 x x

16x = 2 2 2 2 x

2 2 x = 4x

Find the GCF.

The GCF of 8x3, 4x2, and 16x is 4x.

Write terms as products using the GCF as a factor.


Multiply to check your answer.The product is the original

polynomials.

Check

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Example 1C: Factoring by Using the GCFFactor each polynomial. Check your answer.

–14x – 12x2

– 1(14x + 12x2) Both coefficients are negative. Factor out –1.

Find the GCF.

The GCF of 14x and 12x2 is 2x.

–1[7(2x) + 6x(2x)]

–1[2x(7 + 6x)]

–2x(7 + 6x)

Write each term as a product using the GCF.


14x = 2 7 x12x2 = 2 2 3 x x

2 x = 2x

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When you factor out –1 as the first step, be sure to include it in all the other steps as well.

Caution!

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Example 1D: Factoring by Using the GCF

Factor each polynomial. Check your answer.

3x3 + 2x2 – 10

10 = 2 5

Find the GCF.


The polynomial cannot be factored further.

3x3 + 2x2 – 10

3x3 = 3 x x x

2x2 = 2 x x

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5b + 9b3

5b = 5 b

9b = 3 3 b b b

b

5(b) + 9b2(b)

b(5 + 9b2)

b(5 + 9b2)Check

5b + 9b3

Find the GCF.

The GCF of 5b and 9b3 is b.

Multiply to check your answer.

The product is the original polynomial.



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9d2 – 82

Find the GCF.


The polynomial cannot be factored further.

9d2 – 82

9d2 = 3 3 d d

82 = 2 2 2 2 2 2

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Check It Out! Example 1c


–18y3 – 7y2 – 1(18y3 + 7y2) Both coefficients are negative.

Factor out –1.

Find the GCF.

The GCF of 18y3 and 7y2 is y2.

18y3 = 2 3 3 y y y

7y2 = 7 y y

y y = y2

Write each term as a product using the GCF.

Use the Distributive Property to factor out the GCF..

–1[18y(y2) + 7(y2)]

–1[y2(18y + 7)]

–y2(18y + 7)

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Check It Out! Example 1d


8x4 + 4x3 – 2x2

8x4 = 2 2 2 x x x x4x3 = 2 2 x x x2x2 = 2 x x

2 x x = 2x2

4x2(2x2) + 2x(2x2) –1(2x2)

2x2(4x2 + 2x – 1)

Check 2x2(4x2 + 2x – 1)

8x4 + 4x3 – 2x2

The GCF of 8x4, 4x3 and –2x2 is 2x2.

Multiply to check your answer.




Find the GCF.

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To write expressions for the length and width of a rectangle with area expressed by a polynomial, you need to write the polynomial as a product. You can write a polynomial as a product by factoring it.

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Example 2: Application

The area of a court for the game squash is 9x2 + 6x m2. Factor this polynomial to find possible expressions for the dimensions of the squash court.

A = 9x2 + 6x

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

= 3x(3x + 2)

Possible expressions for the dimensions of the squash court are 3x m and (3x + 2) m.

The GCF of 9x2 and 6x is 3x.

Write each term as a product using the GCF as a factor.


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Sometimes the GCF of terms is a binomial. This GCF is called a common binomial factor. You factor out a common binomial factor the same way you factor out a monomial factor.

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Example 3: Factoring Out a Common Binomial Factor

Factor each expression.

A. 5(x + 2) + 3x(x + 2)

5(x + 2) + 3x(x + 2)

(x + 2)(5 + 3x)

The terms have a common binomial factor of (x + 2).

Factor out (x + 2).

B. –2b(b2 + 1)+ (b2 + 1)

–2b(b2 + 1) + (b2 + 1)

–2b(b2 + 1) + 1(b2 + 1)

(b2 + 1)(–2b + 1)

The terms have a common binomial factor of (b2 + 1).

(b2 + 1) = 1(b2 + 1)

Factor out (b2 + 1).

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a. 4s(s + 6) – 5(s + 6)

4s(s + 6) – 5(s + 6) The terms have a common binomial factor of (s + 6).

(4s – 5)(s + 6) Factor out (s + 6).

b. 7x(2x + 3) + (2x + 3)

7x(2x + 3) + (2x + 3)

7x(2x + 3) + 1(2x + 3)

(2x + 3)(7x + 1)

The terms have a common binomial factor of (2x + 3).

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

Factor out (2x + 3).

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You may be able to factor a polynomial by grouping. When a polynomial has four terms, you can make two groups and factor out the GCF from each group.

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Example 4A: Factoring by Grouping

Factor each polynomial by grouping. Check your answer.

6h4 – 4h3 + 12h – 8

(6h4 – 4h3) + (12h – 8)

2h3(3h – 2) + 4(3h – 2)

2h3(3h – 2) + 4(3h – 2)

(3h – 2)(2h3 + 4)

Group terms that have a common number or variable as a factor.

Factor out the GCF of each group.

(3h – 2) is another common factor.

Factor out (3h – 2).

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Example 4A Continued


Check (3h – 2)(2h3 + 4) Multiply to check your solution.

3h(2h3) + 3h(4) – 2(2h3) – 2(4)

6h4 + 12h – 4h3 – 8


6h4 – 4h3 + 12h – 8

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Example 4B: Factoring by Grouping


5y4 – 15y3 + y2 – 3y

(5y4 – 15y3) + (y2 – 3y)

5y3(y – 3) + y(y – 3)

5y3(y – 3) + y(y – 3)

(y – 3)(5y3 + y)

Group terms.


(y – 3) is a common factor.

Factor out (y – 3).

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6b3 + 8b2 + 9b + 12

(6b3 + 8b2) + (9b + 12)

2b2(3b + 4) + 3(3b + 4)

2b2(3b + 4) + 3(3b + 4)

(3b + 4)(2b2 + 3)

Group terms.


(3b + 4) is a common factor.

Factor out (3b + 4).

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4r3 + 24r + r2 + 6

(4r3 + 24r) + (r2 + 6)

4r(r2 + 6) + 1(r2 + 6)

4r(r2 + 6) + 1(r2 + 6)

(r2 + 6)(4r + 1)

Group terms.


(r2 + 6) is a common factor.

Factor out (r2 + 6).

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If two quantities are opposites, their sum is 0.

(5 – x) + (x – 5)

5 – x + x – 5

– x + x + 5 – 5

0 + 0

0

Helpful Hint

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Recognizing opposite binomials can help you factor polynomials. The binomials (5 – x) and (x – 5) are opposites. Notice (5 – x) can be written as –1(x – 5).

–1(x – 5) = (–1)(x) + (–1)(–5)

= –x + 5

= 5 – x

So, (5 – x) = –1(x – 5)

Distributive Property.

Simplify.

Commutative Property of Addition.

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Example 5: Factoring with Opposites

Factor 2x3 – 12x2 + 18 – 3x

2x3 – 12x2 + 18 – 3x

(2x3 – 12x2) + (18 – 3x)

2x2(x – 6) + 3(6 – x)

2x2(x – 6) + 3(–1)(x – 6)

2x2(x – 6) – 3(x – 6)

(x – 6)(2x2 – 3)

Group terms.


Simplify. (x – 6) is a common factor.

Factor out (x – 6).

Write (6 – x) as –1(x – 6).

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15x2 – 10x3 + 8x – 12

(15x2 – 10x3) + (8x – 12)

5x2(3 – 2x) + 4(2x – 3)

5x2(3 – 2x) + 4(–1)(3 – 2x)

5x2(3 – 2x) – 4(3 – 2x)

(3 – 2x)(5x2 – 4)

Group terms.Factor out the GCF of

each group.

Simplify. (3 – 2x) is a common factor.

Factor out (3 – 2x).

Write (2x – 3) as –1(3 – 2x).

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8-3 Factoring x2 + bx + c

5 Minute Warm-Up

Factor each polynomial.

1. 16x + 20x3

2. 4m4 – 12m2 + 8m

3. 7k(k – 3) + 4(k – 3)

4. 3y(2y + 3) – 5(2y + 3)

5. 2x3 + x2 – 6x – 3

6. 7p4 – 2p3 + 63p – 18

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Factor quadratic trinomials of the form x2 + bx + c.

Objective

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In Chapter 7, you learned how to multiply two binomials using the Distributive Property or the FOIL method. In this lesson, you will learn how to factor a trinomial into two binominals.

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(x + 2)(x + 5) = x2 + 7x + 10

You can use this fact to factor a trinomial into its binomial factors. Look for two numbers that are factors of the constant term in the trinomial. Write two binomials with those numbers, and then multiply to see if you are correct.

Notice that when you multiply (x + 2)(x + 5), the constant term in the trinomial is the product of the constants in the binomials.

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Example 1A: Factoring Trinomials by Guess and Check

Factor x2 + 15x + 36 by guess and check.

( + )( + )

(x + )(x + )

Write two sets of parentheses.

The first term is x2, so the variable terms have a coefficient of 1.

The constant term in the trinomial is 36.

(x + 1)(x + 36) = x2 + 37x + 36

(x + 2)(x + 18) = x2 + 20x + 36

(x + 3)(x + 12) = x2 + 15x + 36

Try factors of 36 for the constant terms in the binomials.

The factors of x2 + 15x + 36 are (x + 3)(x + 12).

x2 + 15x + 36 = (x + 3)(x + 12)

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When you multiply two binomials, multiply:

First terms

Outer terms

Inner terms

Last terms

Remember!

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Check It Out! Example 1a Factor each trinomial by guess and check.

x2 + 10x + 24

( + )( + )

(x + )(x + )

Write two sets of parentheses.

The first term is x2, so the variable terms have a coefficient of 1.

The constant term in the trinomial is 24.

(x + 1)(x + 24) = x2 + 25x + 24

(x + 2)(x + 12) = x2 + 14x + 24

(x + 3)(x + 8) = x2 + 11x + 24

Try factors of 24 for the constant terms in the binomials.

(x + 4)(x + 6) = x2 + 10x + 24

The factors of x2 + 10x + 24 are (x + 4)(x + 6). x2 + 10x + 24 = (x + 4)(x + 6)

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The guess and check method is usually not the most efficient method of factoring a trinomial. Look at the product of (x + 3) and (x + 4).

(x + 3)(x +4) = x2 + 7x + 12

x2 12

3x4x

The coefficient of the middle term is the sum of 3 and 4. The third term is the product of 3 and 4.

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When c is positive, its factors have the same sign. The sign of b tells you whether the factors are positive or negative. When b is positive, the factors are positive and when b is negative, the factors are negative.

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Example 2A: Factoring x2 + bx + c When c is Positive

x2 + 6x + 5

Factor each trinomial. Check your answer.

(x + )(x + ) b = 6 and c = 5; look for factors of 5 whose sum is 6.

Factors of 5 Sum

1 and 5 6 The factors needed are 1 and 5.

(x + 1)(x + 5)

Check (x + 1)(x + 5) = x2 + x + 5x + 5 Use the FOIL method.The product is the

original polynomial.= x2 + 6x + 5

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Example 2C: Factoring x2 + bx + c When c is Positive


x2 – 8x + 15

b = –8 and c = 15; look for factors of 15 whose sum is –8.

The factors needed are –3 and –5 .

Factors of –15 Sum–1 and –15 –16

–3 and –5 –8

(x – 3)(x – 5)

Check (x – 3)(x – 5 ) = x2 – 3x – 5x + 15 Use the FOIL method.The product is the

original polynomial.= x2 – 8x + 15

(x + )(x + )

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x2 – 5x + 6

(x + )(x+ ) b = –5 and c = 6; look for factors of 6 whose sum is –5.

The factors needed are –2 and –3.

Factors of 6 Sum–1 and –6 –7

–2 and –3 –5

(x – 2)(x – 3)

Check (x – 2)(x – 3) = x2 –2x – 3x + 6 Use the FOIL method.The product is the



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Check It Out! Example 2d

x2 – 13x + 40

(x + )(x+ )

b = –13 and c = 40; look for factors of 40 whose sum is –13.

The factors needed are –5 and –8.

(x – 5)(x – 8)

Check (x – 5)(x – 8) = x2 – 5x – 8x + 40 Use the FOIL method.The product is the


Factors of 40 Sum

–2 and –20 –22 –4 and –10 –14

–5 and –8 –13

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When c is negative, its factors have opposite signs. The sign of b tells you which factor is positive and which is negative. The factor with the greater absolute value has the same sign as b.

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Example 3A: Factoring x2 + bx + c When c is Negative

Factor each trinomial.

x2 + x – 20

(x + )(x + ) b = 1 and c = –20; look for factors of –20 whose sum is 1. The factor with the greater absolute value is positive.

The factors needed are +5 and –4.

Factors of –20 Sum

–1 and 20 19 –2 and 10 8

–4 and 5 1

(x – 4)(x + 5)

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x2 – 3x – 18b = –3 and c = –18; look for

factors of –18 whose sum is –3. The factor with the greater absolute value is negative.Factors of –18 Sum

1 and –18 –17 2 and – 9 – 7

3 and – 6 – 3

The factors needed are 3 and –6.

(x – 6)(x + 3)

Example 3B: Factoring x2 + bx + c When c is Negative

(x + )(x + )

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A polynomial and the factored form of the polynomial are equivalent expressions. When you evaluate these two expressions for the same value of the variable, the results are the same.

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Example 4A: Evaluating Polynomials

Factor y2 + 10y + 21. Show that the original polynomial and the factored form have the same value for y = 0, 1, 2, 3, and 4.

y2 + 10y + 21

(y + )(y + ) b = 10 and c = 21; look for factors of 21 whose sum is 10.

Factors of 21 Sum1 and 21 7

3 and 7 10

The factors needed are 3 and 7.

(y + 3)(y + 7)

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Example 4A Continued

Evaluate the original polynomial and the factored form for y = 0, 1, 2, 3, and 4.

(y + 7)(y + 3)

(0 + 7)(0 + 3) = 21

(1 + 7)(1 + 3) = 32

(2 + 7)(2 + 3) = 45

(3 + 7)(3 + 3) = 60

(4 + 7)(4 + 3) = 77

y

0

1

2

3

4

y2 + 10y + 21

02 + 10(0) + 21 = 21

y

0

1

2

3

4

12 + 10(1) + 21 = 32

22 + 10(2) + 21 = 45

32 + 10(3) + 21 = 60

42 + 10(4) + 21 = 77

The original polynomial and the factored form have the same value for the given values of n.

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Lesson Quiz: Part I


1. x2 – 11x + 30

2. x2 + 10x + 9

3. x2 – 6x – 27

4. x2 + 14x – 32

(x + 16)(x – 2)

(x – 9)(x + 3)

(x + 1)(x + 9)

(x – 5)(x – 6)

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8-4 Factoring ax2 + bx + c

Warm Up

Find each product.

1. (x – 2)(2x + 7)

2. (3y + 4)(2y + 9)

3. (3n – 5)(n – 7)

Factor each trinomial.4. x2 +4x – 325. z2 + 15z + 366. h2 – 17h + 72

6y2 + 35y + 36

2x2 + 3x – 14

3n2 – 26n + 35

(z + 3)(z + 12) (x – 4)(x + 8)

(h – 8)(h – 9)

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Factor quadratic trinomials of the form ax2 + bx + c.

Objective

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In the previous lesson you factored trinomials of the form x2 + bx + c. Now you will factor trinomials of the form ax2 + bx + c, where a ≠ 0.

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When you multiply (3x + 2)(2x + 5), the coefficient of the x2-term is the product of the coefficients of the x-terms. Also, the constant term in the trinomial is the product of the constants in the binomials.

(3x + 2)(2x + 5) = 6x2 + 19x + 10

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To factor a trinomial like ax2 + bx + c into its binomial factors, write two sets of parentheses ( x + )( x + ).

Write two numbers that are factors of a next to the x’s and two numbers that are factors of c in the other blanks. Multiply the binomials to see if you are correct.

(3x + 2)(2x + 5) = 6x2 + 19x + 10

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Example 1: Factoring ax2 + bx + c by Guess and Check

Factor 6x2 + 11x + 4 by guess and check.

( + )( + ) Write two sets of parentheses.

The first term is 6x2, so at least one variable term has a coefficient other than 1.

( x + )( x + )

The coefficient of the x2 term is 6. The constant term in the trinomial is 4.

Try factors of 6 for the coefficients and factors of 4 for the constant terms.

(1x + 4)(6x + 1) = 6x2 + 25x + 4 (1x + 2)(6x + 2) = 6x2 + 14x + 4 (1x + 1)(6x + 4) = 6x2 + 10x + 4

(2x + 4)(3x + 1) = 6x2 + 14x + 4

(3x + 4)(2x + 1) = 6x2 + 11x + 4

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( X + )( x + ) = ax2 + bx + c

So, to factor a2 + bx + c, check the factors of a and the factors of c in the binomials. The sum of the products of the outer and inner terms should be b.

Sum of outer and inner products = b

Product = cProduct = a

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Since you need to check all the factors of a and the factors of c, it may be helpful to make a table. Then check the products of the outer and inner terms to see if the sum is b. You can multiply the binomials to check your answer.

( X + )( x + ) = ax2 + bx + c

Sum of outer and inner products = b

Product = cProduct = a

Holt Algebra 1


Example 2A: Factoring ax2 + bx + c When c is Positive


2x2 + 17x + 21

( x + )( x + )a = 2 and c = 21, Outer + Inner = 17.

(x + 7)(2x + 3)

Factors of 2 Factors of 21 Outer + Inner

1 and 2 1 and 21 1(21) + 2(1) = 23

1 and 2 21 and 1 1(1) + 2(21) = 43 1 and 2 3 and 7 1(7) + 2(3) = 13 1 and 2 7 and 3 1(3) + 2(7) = 17

Check (x + 7)(2x + 3) = 2x2 + 3x + 14x + 21= 2x2 + 17x + 21

Use the Foil method.

Holt Algebra 1


When b is negative and c is positive, the factors of c are both negative.

Remember!

Holt Algebra 1



3x2 – 16x + 16a = 3 and c = 16, Outer + Inner = –16 .

(x – 4)(3x – 4)

Check (x – 4)(3x – 4) = 3x2 – 4x – 12x + 16

= 3x2 – 16x + 16



1 and 3 –1 and –16 1(–16) + 3(–1) = –19 1 and 3 – 2 and – 8 1( – 8) + 3(–2) = –14 1 and 3 – 4 and – 4 1( – 4) + 3(– 4)= –16

( x + )( x + )

Example 2B: Factoring ax2 + bx + c When c is Positive

Holt Algebra 1




9x2 – 15x + 4 a = 9 and c = 4, Outer + Inner = –15.


3 and 3 –1 and – 4 3(–4) + 3(–1) = –15 3 and 3 – 2 and – 2 3(–2) + 3(–2) = –12 3 and 3 – 4 and – 1 3(–1) + 3(– 4)= –15

(3x – 4)(3x – 1)

Check (3x – 4)(3x – 1) = 9x2 – 3x – 12x + 4

= 9x2 – 15x + 4


( x + )( x + )

Holt Algebra 1



3x2 + 13x + 12 a = 3 and c = 12, Outer + Inner = 13.


1 and 3 1 and 12 1(12) + 3(1) = 15 1 and 3 2 and 6 1(6) + 3(2) = 12 1 and 3 3 and 4 1(4) + 3(3) = 13

(x + 3)(3x + 4)

Check (x + 3)(3x + 4) = 3x2 + 4x + 9x + 12

= 3x2 + 13x + 12



( x + )( x + )

Holt Algebra 1


When c is negative, one factor of c will be positive and the other factor will be negative. Only some of the factors are shown in the examples, but you may need to check all of the possibilities.

Holt Algebra 1


Example 3A: Factoring ax2 + bx + c When c is Negative


3n2 + 11n – 4

( y + )( y+ )a = 3 and c = – 4, Outer + Inner = 11 .

(n + 4)(3n – 1)Check (n + 4)(3n – 1) = 3n2 – n + 12n – 4

= 3n2 + 11n – 4



1 and 3 –1 and 4 1(4) + 3(–1) = 1 1 and 3 –2 and 2 1(2) + 3(–2) = – 4 1 and 3 –4 and 1 1(1) + 3(–4) = –11

1 and 3 4 and –1 1(–1) + 3(4) = 11

Holt Algebra 1



2x2 + 9x – 18

( x + )( x+ )a = 2 and c = –18, Outer + Inner = 9 .

Factors of 2 Factors of – 18 Outer + Inner

1 and 2 18 and –1 1(– 1) + 2(18) = 35 1 and 2 9 and –2 1(– 2) + 2(9) = 16 1 and 2 6 and –3 1(– 3) + 2(6) = 9

(x + 6)(2x – 3)

Check (x + 6)(2x – 3) = 2x2 – 3x + 12x – 18

= 2x2 + 9x – 18


Example 3B: Factoring ax2 + bx + c When c is Negative

Holt Algebra 1



4x2 – 15x – 4

( x + )( x+ )a = 4 and c = –4,Outer + Inner = –15.

Factors of 4 Factors of – 4 Outer + Inner

1 and 4 –1 and 4 1(4) – 1(4) = 0 1 and 4 –2 and 2 1(2) – 2(4) = –6 1 and 4 –4 and 1 1(1) – 4(4) = –15

(x – 4)(4x + 1) Use the Foil method.

Check (x – 4)(4x + 1) = 4x2 + x – 16x – 4

= 4x2 – 15x – 4

Example 3C: Factoring ax2 + bx + c When c is Negative

Holt Algebra 1


When the leading coefficient is negative, factor out –1 from each term before using other factoring methods.

Holt Algebra 1


When you factor out –1 in an early step, you must carry it through the rest of the steps.

Caution

Holt Algebra 1


Example 4A: Factoring ax2 + bx + c When a is Negative

Factor –2x2 – 5x – 3.

–1(2x2 + 5x + 3)

–1( x + )( x+ )

Factor out –1. a = 2 and c = 3; Outer + Inner = 5


1 and 2 3 and 1 1(1) + 3(2) = 7 1 and 2 1 and 3 1(3) + 1(2) = 5

–1(x + 1)(2x + 3)

(x + 1)(2x + 3)

Holt Algebra 1




–6x2 – 17x – 12

–1(6x2 + 17x + 12)

–1( x + )( x+ )

Factor out –1. a = 6 and c = 12; Outer + Inner = 17


2 and 3 4 and 3 2(3) + 3(4) = 18 2 and 3 3 and 4 2(4) + 3(3) = 17

(2x + 3)(3x + 4) –1(2x + 3)(3x + 4)

Holt Algebra 1




–3x2 – 17x – 10

–1(3x2 + 17x + 10)

–1( x + )( x+ )

Factor out –1. a = 3 and c = 10; Outer + Inner = 17)


1 and 3 2 and 5 1(5) + 3(2) = 11 1 and 3 5 and 2 1(2) + 3(5) = 17

(3x + 2)(x + 5) –1(3x + 2)(x + 5)

Holt Algebra 1

8-5 Factoring Special Products

5 Minute Warm-Up


1. 5x2 + 17x + 6

2. 2x2 + 5x – 12

3. 6x2 – 23x + 7

4. –4x2 + 11x + 20

5. –2x2 + 7x – 3

6. 8x2 + 27x + 9

(–x + 4)(4x + 5)

(3x – 1)(2x – 7)

(2x– 3)(x + 4)

(5x + 2)(x + 3)

(–2x + 1)(x – 3)

(8x + 3)(x + 3)

Holt Algebra 1


5 Minute Warm-Up


1. 40p2 – 10p + 30

2. 5g5 – 10g3 - 15g

3. 2x(x-4) + 9(x-4)

4. 10m3 + 15m2 – 2m - 3

5. x2 + 14x - 120

6. 6x2 – 19x + 15

Holt Algebra 1


Factor perfect-square trinomials.

Factor the difference of two squares.

Objectives

Holt Algebra 1


A trinomial is a perfect square if:

• The first and last terms are perfect squares.

• The middle term is two times one factor from the first term and one factor from the last term.

9x2 + 12x + 4

3x 3x 2(3x 2) 2 2• ••

Holt Algebra 1


Holt Algebra 1


Example 1A: Recognizing and Factoring Perfect-Square Trinomials

Determine whether each trinomial is a perfect square. If so, factor. If not explain.

9x2 – 15x + 64

9x2 – 15x + 64

2(3x 8) ≠ –15x.

9x2 – 15x + 64 is not a perfect-square trinomial because –15x ≠ 2(3x 8).

8 83x 3x 2(3x 8)

Holt Algebra 1


Example 1B: Recognizing and Factoring Perfect-Square Trinomials


81x2 + 90x + 25

81x2 + 90x + 25

The trinomial is a perfect square. Factor.

5 59x 9x 2(9x 5)● ●●

Holt Algebra 1


Example 1B Continued


Method 2 Use the rule.

81x2 + 90x + 25 a = 9x, b = 5

(9x)2 + 2(9x)(5) + 52

(9x + 5)2

Write the trinomial as a2 + 2ab + b2.

Write the trinomial as (a + b)2.

Holt Algebra 1


Example 1C: Recognizing and Factoring Perfect-Square Trinomials


36x2 – 10x + 14

The trinomial is not a perfect-square because 14 is not a perfect square.

36x2 – 10x + 14

36x2 – 10x + 14 is not a perfect-square trinomial.

Holt Algebra 1


A rectangular piece of cloth must be cut to make a tablecloth. The area needed is (16x2 – 24x + 9) in2. The dimensions of the cloth are of the form cx – d, where c and d are whole numbers. Find an expression for the perimeter of the cloth. Find the perimeter when x = 11 inches.

Example 2: Problem-Solving Application

Holt Algebra 1


2 Make a Plan

The formula for the area of a square is area = (side)2.

Example 2 Continued

Factor 16x2 – 24x + 9 to find the side length of the tablecloth. Write a formula for the perimeter of the park, and evaluate the expression for x = 11.

Holt Algebra 1


Solve3

16x2 – 24x + 9

(4x)2 – 2(4x)(3) + 32

(4x – 3)2

16x2 – 24x + 9 = (4x – 3)(4x – 3)

a = 4x, b = 3

Write the trinomial as a2 – 2ab + b2.

Write the trinomial as (a + b)2.

The side length of the tablecloth is (4x – 3) in. and (4x – 3) in.

Example 2 Continued

Holt Algebra 1


Write a formula for the perimeter of the tablecloth.

P = 4s

= 4(4x – 3)

= 16x – 12

An expression for the perimeter of the tablecloth in inches is 16x – 12.

Write the formula for the perimeter of a square.

Substitute the side length for s.

Distribute 4.

Example 2 Continued

Holt Algebra 1


Evaluate the expression when x = 11.

P = 16x – 12

= 16(11) – 12

= 164

When x = 11 in. the perimeter of the tablecloth is 164 in.

Substitute 11 for x.

Example 2 Continued

Holt Algebra 1


In Chapter 7 you learned that the difference of two squares has the form a2 – b2. The difference of two squares can be written as the product (a + b)(a – b). You can use this pattern to factor some polynomials.

A polynomial is a difference of two squares if:

• There are two terms, one subtracted from the other.

• Both terms are perfect squares.

4x2 – 9

2x 2x 3 3

Holt Algebra 1


Holt Algebra 1


Recognize a difference of two squares: the coefficients of variable terms are perfect squares, powers on variable terms are even, and constants are perfect squares.

Reading Math

Holt Algebra 1


Example 3A: Recognizing and Factoring the Difference of Two Squares

Determine whether each binomial is a difference of two squares. If so, factor. If not, explain.

3p2 – 9q4

3p2 – 9q4

3q2 3q2 3p2 is not a perfect square.

3p2 – 9q4 is not the difference of two squares because 3p2 is not a perfect square.

Holt Algebra 1


Example 3B: Recognizing and Factoring the Difference of Two Squares


100x2 – 4y2

Write the polynomial as (a + b)(a – b).

a = 10x, b = 2y

The polynomial is a difference of two squares.

100x2 – 4y2

2y 2y10x 10x

(10x)2 – (2y)2 (10x + 2y)(10x – 2y)

100x2 – 4y2 = (10x + 2y)(10x – 2y)

Holt Algebra 1


Example 3C: Recognizing and Factoring the Difference of Two Squares


x4 – 25y6


a = x2, b = 5y3


(x2)2 – (5y3)2

(x2 + 5y3)(x2 – 5y3)

x4 – 25y6 = (x2 + 5y3)(x2 – 5y3)

5y3 5y3x2 x2

x4 – 25y6

Holt Algebra 1




1 – 4x2


a = 1, b = 2x


(1)2 – (2x)2 (1 + 2x)(1 – 2x)

1 – 4x2 = (1 + 2x)(1 – 2x)

2x 2x1 1

1 – 4x2

Holt Algebra 1




p8 – 49q6


a = p4, b = 7q3


(p4)2 – (7q3)2 (p4 + 7q3)(p4 – 7q3)

p8 – 49q6 = (p4 + 7q3)(p4 – 7q3)

7q3 7q3●p4 p4●

p8 – 49q6

Holt Algebra 1




16x2 – 4y5

4x 4x 4y5 is not a perfect square.

16x2 – 4y5 is not the difference of two squares because 4y5 is not a perfect square.

16x2 – 4y5

Holt Algebra 1

8-6 Choosing a Factoring Method

5 Minute Warm-up

Determine whether each trinomial is a perfect square. If so factor. If not, explain.

1. 64x2 – 40x + 25

2. 121x2 – 44x + 4

3. 49x2 + 140x + 100

4. A square fence will be built around a garden with an area of (49x2 + 56x + 16) ft2. The dimensions of the garden are cx + d, where c and d are

whole numbers. Find an expression for the perimeter when x = 5.

P = 28x + 16; 156 ft

(7x2 + 10)2

(11x – 2)2

Not a perfect-square trinomial because –40x ≠ 2(8x 5).

Holt Algebra 1


Choose an appropriate method for factoring a polynomial.

Combine methods for factoring a polynomial.

Objectives

Holt Algebra 1


Recall that a polynomial is in its fully factored form when it is written as a product that cannot be factored further.

Holt Algebra 1


Example 1: Determining Whether a Polynomial is Completely Factored

Tell whether each polynomial is completely factored. If not factor it.

A. 3x2(6x – 4)

3x2(6x – 4)

Neither x2 +1 nor x – 5 can be factored further.

6x2(3x – 2)

B. (x2 + 1)(x – 5)

(x2 + 1)(x – 5)

6x – 4 can be further factored.

Factor out 2, the GCF of 6x and – 4.

6x2(3x – 2) is completely factored.

(x2 + 1)(x – 5) is completely factored.

Holt Algebra 1


x2 + 4 is a sum of squares, and cannot be factored.

Caution

Holt Algebra 1



Tell whether the polynomial is completely factored. If not, factor it.

A. 5x2(x – 1)

5x2(x – 1) Neither 5x2 nor x – 1 can be factored further.

B. (4x + 4)(x + 1)

Factor out 4, the GCF of 4x and 4.

4x + 4 can be further factored.

5x2(x – 1) is completely factored.

4(x + 1)2 is completely factored.

(4x + 4)(x + 1)

4(x + 1)(x + 1)

Holt Algebra 1


To factor a polynomial completely, you may need to use more than one factoring method. Use the steps below to factor a polynomial completely.

Holt Algebra 1


Example 2A: Factoring by GCF and Recognizing Patterns

Factor 10x2 + 48x + 32 completely. Check your answer.

10x2 + 48x + 32

2(5x2 + 24x + 16)

2(5x + 4)(x + 4)

Factor out the GCF.

Check 2(5x + 4)(x + 4) = 2(5x2 + 20x + 4x + 16)

= 10x2 + 40x + 8x + 32

= 10x2 + 48x + 32

Factor remaining trinomial.

Holt Algebra 1


Example 2B: Factoring by GCF and Recognizing Patterns

Factor 8x6y2 – 18x2y2 completely. Check your answer.

8x6y2 – 18x2y2

2x2y2(4x4 – 9)

Factor out the GCF. 4x4 – 9 is a perfect-square trinomial of the form a2 – b2.

2x2y2(2x2 – 3)(2x2 + 3) a = 2x, b = 3

Check 2x2y2(2x2 – 3)(2x2 + 3) = 2x2y2(4x4 – 9)

= 8x6y2 – 18x2y2

Holt Algebra 1



Factor each polynomial completely. Check your answer.

4x3 + 16x2 + 16x

4x3 + 16x2 + 16x

4x(x2 + 4x + 4)

4x(x + 2)2

Factor out the GCF. x2 + 4x + 4 is a perfect-square trinomial of the form a2 + 2ab + b2.

a = x, b = 2

Check 4x(x + 2)2 = 4x(x2 + 2x + 2x + 4)

= 4x(x2 + 4x + 4)

= 4x3 + 16x2 + 16x

Holt Algebra 1




2x2y – 2y3

2y(x + y)(x – y)

Factor out the GCF. 2y(x2 – y2) is a perfect-square trinomial of the form a2 – b2.

a = x, b = y

Check 2y(x + y)(x – y) = 2y(x2 + xy – xy – y2)

= 2x2y – 2y3

2x2y – 2y3

2y(x2 – y2)

= 2x2y +2xy2 – 2xy2 – 2y3

Holt Algebra 1


If none of the factoring methods work, the polynomial is said to be unfactorable.

For a polynomial of the form ax2 + bx + c, if there are no numbers whose sum is b and whose product is ac, then the polynomial is unfactorable.

Helpful Hint

Holt Algebra 1


Example 3A: Factoring by Multiple Methods

Factor each polynomial completely.

9x2 + 3x – 2

9x2 + 3x – 2 ( x + )( x + )

The GCF is 1 and there is no pattern.

a = 9 and c = –2; Outer + Inner = 3


1 and 9 1 and –2 1(–2) + 1(9) = 7

3 and 3 1 and –2 3(–2) + 1(3) = –3 3 and 3 –1 and 2 3(2) + 3(–1) = 3

(3x – 1)(3x + 2)

Holt Algebra 1


Example 3B: Factoring by Multiple Methods


12b3 + 48b2 + 48b

(x + )(x + )

The GCF is 12b; (b2 + 4b + 4) is a perfect-square trinomial in the form of a2 + 2ab + b2.

a = 2 and c = 2

12b(b2 + 4b + 4)

12b(b + 2)(b + 2)

12b(b + 2)2

Factors of 4 Sum

1 and 4 52 and 2 4

Holt Algebra 1


Example 3C: Factoring by Multiple Methods


4y2 + 12y – 72

4(y2 + 3y – 18) Factor out the GCF. There is no

pattern. b = 3 and c = –18; look for factors of –18 whose sum is 3. (y + )(y + )

Factors of –18 Sum

–1 and 18 17 –2 and 9 7

–3 and 6 3

4(y – 3)(y + 6)

The factors needed are –3 and 6

Holt Algebra 1


Example 3D: Factoring by Multiple Methods.


(x4 – x2)

x2(x2 – 1) Factor out the GCF.

x2(x + 1)(x – 1) x2 – 1 is a difference of two squares.

Holt Algebra 1




2p5 + 10p4 – 12p3

2p3(p2 + 5p – 6) Factor out the GCF. There is no pattern. b = 5 and c = –6; look for factors of –6 whose sum is 5.

(p + )(p + )

Factors of – 6 Sum

– 1 and 6 5

2p3(p + 6)(p – 1)

The factors needed are –1 and 6

Holt Algebra 1




Factor out the GCF. There is no pattern.3q4(3q2 + 10q + 8)

9q6 + 30q5 + 24q4

a = 3 and c = 8; Outer + Inner = 10

( q + )( q + )


3 and 1 1 and 8 3(8) + 1(1) = 25

3 and 1 2 and 4 3(4) + 1(2) = 14 3 and 1 4 and 2 3(2) + 1(4) = 10

3q4(3q + 4)(q + 2)

Holt Algebra 1


Holt Algebra 1


Tell whether the polynomial is completely factored. If not, factor it.1. (x + 3)(5x + 10) 2. 3x2(x2 + 9)

Lesson Quiz

(x + 4)(x2 + 3)

completely factoredno; 5(x+ 3)(x + 2)

4(x + 6)(x – 2)

5. 18x2 – 3x – 3 6. 18x2 – 50y2

7. 5x – 20x3 + 7 – 28x2

3(3x + 1)(2x – 1) 2(3x + 5y)(3x – 5y)

(1 + 2x)(1 – 2x)(5x + 7)


3. x3 + 4x2 + 3x + 12 4. 4x2 + 16x – 48

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