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Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping because Study Guide and Review - Chapter 8

Chapter 8 Algebra 1 p 1 Barr 2016 Chapter 8: Factoring Polynomials Quiz 8 4 Q u a d ra tics in th e fo rm a x 2 +bx+c 8 5 S p ecia l C a ses Quiz R eview

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Chapter 8 KEYWORD: MA7 ChProj Factoring Polynomials 8A Factoring Methods 8-1 Factors and Greatest examples, and answers to your questions Try This This problem will prepare you for the Multi-Step Test Prep on page 556

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Name Answer Key Date Name Date Class CHAPTER Chapter Test Review 8 Form A 1 Write the prime factorization of 36 7 Factor a* Factor 5 30y3 – 50y 12 Write the polynomial modeled by this geometric diagram and then factor x2

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Simplify expressions involving the quotient of monomials (with 8-1 (with 8-1 Standardized Test Practice (pp mials to factoring polynomials in Chapter 9 Answers 2a (5m)2 25m2 2b The power of a product is the product of the powers

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1 avr 2020 · b : I can factor polynomial expressions using grouping #1 Consider this expression: a Simplify the expression Write the expression in standard

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Solve each equation by factoring and using the zero product property Be able to check the answers in the original equations 27) p2 + 8p + 7 = 0 28)

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State whether each sentence is true or false. If false, replace the underlined phrase or expression to

make a true sentence. x2 + 5x + 6 is an example of a prime polynomial.

The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients

is called a prime polynomial. The polynomial x2 + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The

polynomial x2 + 5x + 7 is an example of a prime polynomial. (x + 5)(x 5) is the factorization of a difference of squares.

The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x 5) is

the factorization of a difference of squares. The statement is true. (x + 5)(x 2) is the factored form of x2 3x 10. (x 5)(x + 2) is the factored form of x2 3x 10. So, the statement is false. Expressions with four or more unlike terms can sometimes be factored by grouping.

Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping because

terms are put into groups and then factored. So, the statement is true. The Zero Product Property states that if ab = 1, then a or b is 1.

The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b

= 0, or a and b are zero. x2 12x + 36 is an example of a perfect square trinomial. Perfect square trinomials are trinomials that are the squares of binomials.

So, the statement is true.

x2 16 is an example of a perfect square trinomial. The statement is false. Perfect square trinomials are trinomials that are the squares of binomials. x2 16 is the product of a sum and a difference. So, x2 16 an example of a difference of squares.

4x2 2x + 7 is a polynomial of degree 2.

true

The leading coefficient of 1 + 6a + 9a2 is 1.

The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient

of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.

The FOIL method is used to multiply two trinomials. false; binomials

FOIL Method:

To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L

and the Last terms.

Write each polynomial in standard form.

x + 2 + 3x2 The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x2 + x + 2. 1 x4 The greatest degree is 4. Therefore, the polynomial can be rewritten as x4 + 1.

2 + 3x + x2

The greatest degree is 2. Therefore, the polynomial can be rewritten as x2 + 3x + 2.

3x5 2 + 6x 2x2 + x3

The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x5 + x3 2x2 + 6x 2.

Find each sum or difference.

(x3 + 2) + (3x3 5) a2 + 5a 3 (2a2 4a + 3) (4x 3x2 + 5) + (2x2 5x + 1) Jean is framing a painting that is a rectangle. What is the perimeter of the frame?

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

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

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

2(4w + w2) 6 = 2w(w 4) + 10

Find the area of the rectangle.

The area of the rectangle is 3x3 + 3x2 21x.

Find each product.

(x 3)(x + 7) (3a 2)(6a + 5) (3r 7t)(2r + 5t) (2x + 5)(5x + 2) The parking lot shown is to be paved. What is the area to be paved?

The area to be paved is 10x2 + 7x 12 units2.

Find each product.

(x + 5)(x 5) (3x 2)2 (5x + 4)2 (2x 3)(2x + 3) (2r + 5t)2 (3m 2)(3m + 2) Write an expression to represent the area of the shaded region.

Find the area of the larger rectangle.

Find the area of the smaller rectangle.

To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger

rectangle.

The area of the shaded region is 3x2 21.

Use the Distributive Property to factor each polynomial.

12x + 24y

Factor

12x + 24y = 12(x + 2y)

14x2y 21xy + 35xy2

Factor

The greatest common factor of each term is 7xy.

14x2y 21xy + 35xy2 = 7xy(2x 3 + 5y)

8xy 16x3y + 10y

Factor.

The greatest common factor of each term is 2y.

8xy 16x3y + 10y = 2y(4x 8x3 + 5)

a2 4ac + ab 4bc

Factor by grouping.

2x2 3xz 2xy + 3yz

24am 9an + 40bm 15bn

Solve each equation. Check your solutions.

6x2 = 12x

Factor the trinomial using the Zero Product Property. The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation. and

The solutions are 0 and 2.

x2 = 3x Factor the trinomial using the Zero Product Property. The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation. and

The solutions are 0 and 3.

3x2 = 5x

Factor the trinomial using the Zero Product Property. The roots are 0 and . Check by substituting 0 and in for x in the original equation. and

The solutions are 0 and .

x(3x 6) = 0 Factor the trinomial using the Zero Product Property. x(3x 6) = 0 x The roots are 0 and 2. Check by substituting 0 and 2 in for x in the original equation. and

The solutions are 0 and 2.

The area of the rectangle shown is x3 2x2 + 5x square units. What is the length?

The area of the rectangle is x3 2x2 + 5x or x(x2 2x + 5). Area is found by multiplying the length by the width.

Because the width is x, the length must be x2 2x + 5. Factor each trinomial. Confirm your answers using a graphing calculator. x2 8x + 15

In this trinomial, b = 8 and c = 15, so m + p is negative and mp is positive. Therefore, m and p must both be

negative. List the negative factors of 15, and look for the pair of factors with a sum of 8.

The correct factors are 3 and 5.

Check using a Graphing calculator.

[10, 10] scl: 1 by [10, 10] scl: 1 Factors of 15 Sum of 8

1, 15 16

3, 5 8

x2 + 9x + 20

In this trinomial, b = 9 and c = 20, so m + p is positive and mp is positive. Therefore, m and p must both be positive.

List the positive factors of 20, and look for the pair of factors with a sum of 9.

The correct factors are 4 and 5.

Check using a Graphing calculator.

[10, 10] scl: 1 by [10, 10] scl: 1 Factors of 20 Sum of 9

1, 20 21

2, 10 12

4, 5 9

x2 5x 6

In this trinomial, b = 5 and c = 6, so m + p is negative and mp is negative. Therefore, m and p must have different

signs. List the factors of 6, and look for the pair of factors with a sum of 5.

The correct factors are 1 and 6.

Check using a Graphing calculator.

[10, 10] scl: 1 by [12, 8] scl: 1 Factors of 6 Sum of 5

1, 6 5

2, 3 1

x2 + 3x 18

In this trinomial, b = 3 and c = 18, so m + p is positive and mp is negative. Therefore, m and p must have different

signs. List the factors of 18, and look for the pair of factors with a sum of 3.

The correct factors are 3 and 6.

Check using a Graphing calculator.

[10, 10] scl: 1 by [14, 6] scl: 1 Factors of 18 Sum of 3

1, 18 17

2, 9 7

3, 6 3

Solve each equation. Check your solutions.

x2 + 5x 50 = 0 The roots are 10 and 5. Check by substituting 10 and 5 in for x in the original equation. and

The solutions are 10 and 5.

x2 6x + 8 = 0 The roots are 2 and 4. Check by substituting 2 and 4 in for x in the original equation. and

The solutions are 2 and 4.

x2 + 12x + 32 = 0 The roots are 8 and 4. Check by substituting 8 and 4 in for x in the original equation. and

The solutions are 8 and 4.

x2 2x 48 = 0 The roots are 6 and 8. Check by substituting 6 and 8 in for x in the original equation. and

The solutions are 6 and 8.

x2 + 11x + 10 = 0 The roots are 10 and 1. Check by substituting 10 and 1 in for x in the original equation. and

The solutions are 10 and 1.

An artist is working on a painting that is 3 inches longer than it is wide. The area of the painting is 154 square

inches. What is the length of the painting? Let x = the width of the painting. Then, x + 3 = the length of the painting.

Because a painting cannot have a negative dimension, the width is 11 inches and the length is 11 + 3, or 14 inches.

Factor each trinomial, if possible. If the trinomial cannot be factored, write prime.

12x2 + 22x 14

In this trinomial, a = 12, b = 22 and c = 14, so m + p is positive and mp is negative. Therefore, m and p must have

different signs. List the factors of 12(14) or 168 and identify the factors with a sum of 22.

The correct factors are 6 and 28.

So, 12x2 + 22x 14 = 2(2x 1)(3x + 7). Factors of 168

2, 84 82

3, 56 53

4, 42 38

6, 28 22

2y2 9y + 3

In this trinomial, a = 2, b = 9 and c = 3, so m + p is negative and mp is positive. Therefore, m and p must both be

negative.

2(3) = 6

There are no factors of 6 with a sum of 9. So, this trinomial is prime.

3x2 6x 45

In this trinomial, a = 3, b = 6 and c = 45, so m + p is negative and mp is negative. Therefore, m and p must have

different signs. List the factors of 3(45) or 135 and identify the factors with a sum of 6.

The correct factors are 15 and 9.

So, 3x2 6x 45 = 3(x 5)(x + 3). Factors of 135 Sum

1, 135 134

3, 45 42

5, 27 22

9, 15 6

2a2 + 13a 24

In this trinomial, a = 2, b = 13 and c = 24, so m + p is positive and mp is negative. Therefore, m and p must have

different signs. List the factors of 2(24) or 48 and identify the factors with a sum of 13.

The correct factors are 3 and 16.

So, 2a2 + 13a 24 = (2a 3)(a + 8). Factors of 48

1, 48 47

2, 24 22

3, 16 13

4, 12 8

6, 8 2

Solve each equation. Confirm your answers using a graphing calculator.

40x2 + 2x = 24

The roots are or

Confirm the roots using a graphing calculator. Let Y1 = 40x2 + 2x and Y2 = 24. Use the intersect option from the

CALC

The solutions are .

[5, 5] scl: 1 by [5, 25] scl: 3 [5, 5] scl: 1 by [5, 25] scl: 3

2x2 3x 20 = 0

The roots are or

Confirm the roots using a graphing calculator. Let Y1 = 2x2 3x 20 and Y2 = 0. Use the intersect option from

the CALC

The solutions are

[10, 10] scl: 1 by [15, 5] scl: 1 [10, 10] scl: 1 by [15, 5] scl: 1

16t2 + 36t 8 = 0

The roots are 2 and

Confirm the roots using a graphing calculator. Let Y1 = 16t2 + 36t and Y2 = 0. Use the intersect option from

the CALC

The solutions are 2 and .

[2, 3] scl: 1 by [20, 10] scl: 6 [2, 3] scl: 1 by [20, 10] scl: 6

6x2 7x 5 = 0

The roots are or

Confirm the roots using a graphing calculator. Let Y1 = 6x2 7x 5 and Y2 = 0. Use the intersect option from the

CALC

The solutions are .

[5, 5] scl: 0.5 by [10, 10] scl: 1 [5, 5] scl: 0.5 by [10, 10] scl: 1 The area of the rectangle shown is 6x2 + 11x 7 square units. What is the width of the rectangle? To find the width, factor the area of the rectangle.

In the area trinomial, a = 6, b = 11 and c = 7, so m + p is positive and mp is negative. Therefore, m and p must have

different signs. List the factors of 6(7) or 42 and identify the factors with a sum of 11.

The correct factors are 3 and 14.

So, 6x2 + 11x 7 = (2x 1)(3x + 7). The area of a rectangle is found by multiplying the length by the width.

Because the length of the rectangle is 2x 1, the width must be 3x + 7. Factors of 42 Sum

1, 42 41

2, 21 19

3, 14 11

6, 7 1

Factor each polynomial.

y2 81

64 25x2

16a2 21b2

The number 21 is not a perfect square. So, 16a2 21b2 is prime.

3x2 3

Solve each equation by factoring. Confirm your answers using a graphing calculator.quotesdbs_dbs20.pdfusesText_26

[PDF] [PDF] State whether each sentence is true or false If false , replace the

So, the statement is true.

4x2 2x + 7 is a polynomial of degree 2.

The leading coefficient of 1 + 6a + 9a2 is 1.

FOIL Method:

Write each polynomial in standard form.

2 + 3x + x2

3x5 2 + 6x 2x2 + x3

Find each sum or difference.

The perimeter of the frame is 4x2 + 4x + 8.

Solve each equation.

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

2(4w + w2) 6 = 2w(w 4) + 10

Find the area of the rectangle.

The area of the rectangle is 3x3 + 3x2 21x.

Find each product.

The area to be paved is 10x2 + 7x 12 units2.

Find each product.

Find the area of the larger rectangle.

Find the area of the smaller rectangle.

The area of the shaded region is 3x2 21.

12x + 24y

Factor

12x + 24y = 12(x + 2y)

14x2y 21xy + 35xy2

Factor

The greatest common factor of each term is 7xy.

14x2y 21xy + 35xy2 = 7xy(2x 3 + 5y)

8xy 16x3y + 10y

Factor.

The greatest common factor of each term is 2y.

8xy 16x3y + 10y = 2y(4x 8x3 + 5)

Factor by grouping.

2x2 3xz 2xy + 3yz

24am 9an + 40bm 15bn

Solve each equation. Check your solutions.

6x2 = 12x

The solutions are 0 and 2.

The solutions are 0 and 3.

3x2 = 5x

The solutions are 0 and .

The solutions are 0 and 2.

The correct factors are 3 and 5.

Check using a Graphing calculator.

1, 15 16

3, 5 8

The correct factors are 4 and 5.

Check using a Graphing calculator.

1, 20 21

2, 10 12

4, 5 9

The correct factors are 1 and 6.

Check using a Graphing calculator.

1, 6 5

1, 6 5

2, 3 1

2, 3 1

The correct factors are 3 and 6.

Check using a Graphing calculator.

1, 18 17

1, 18 17

2, 9 7

2, 9 7

3, 6 3

3, 6 3

Solve each equation. Check your solutions.

The solutions are 10 and 5.

The solutions are 2 and 4.

The solutions are 8 and 4.

The solutions are 6 and 8.

The solutions are 10 and 1.

12x2 + 22x 14

The correct factors are 6 and 28.

2, 84 82

2, 84 82

3, 56 53

3, 56 53

4, 42 38

4, 42 38

6, 28 22