1. 5 Minute Warm-Up Directions: Simplify each problem. Write the
answer in standard form.
(4x + 7)(– 2x)
-4x2(3x2 + 2x – 6)
(x + 6)(x + 9)
(-4y + 5)(-7 – 3y)
(-8x3 + x – 9x2 + 2) + (8x2 – 2x + 4)
6. (6x2 – x + 3) – (-2x + x2 – 7)

2. Warm Up 1. 50, 6 2. 105, 7 3. List the factors of 28. no yes
1. 50, , 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.
Tell whether the second number is a factor of the first number no
yes
±1, ±2, ±4, ±7,
±14, ±28
4. 11
prime
5. 98
composite; 49  2

3. Objectives Write the prime factorization of numbers.
Find the GCF of monomials.

4. 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
1 12 
2 6
3 4

5. 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.
Factorizations of 12
1 12 
2 6
3 4

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

7. Example 1: Writing Prime Factorizations
Write the prime factorization of 98.
Method 1 Factor tree
Method 2 Ladder diagram
Choose any two factors
of 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  98 49 7 1 2
98 = 
98 = 
The prime factorization of 98 is 2  7  7 or 2  72.

8. Check It Out! Example 1
Write the prime factorization of each number.
a. 40
b. 33 40  2 5 33 3 11
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.

9. 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.

10. Example 2A: Finding the GCF of Numbers
Find the GCF of each pair of numbers.
100 and 60
Method 1 List the factors.
factors of 100: 1, 2, 4, 5, 10, 20, 25, 50, 100
List all the factors.
factors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
Circle the GCF.
The GCF of 100 and 60 is 20.

11. Example 2B: Finding the GCF of Numbers
Find the GCF of each pair of numbers.
26 and 52
Method 2 Prime factorization.
Write the prime factorization of each number.
26 =  13
52 = 2  2  13
Align the common factors.
2  13 = 26
The GCF of 26 and 52 is 26.

12. You can also find the GCF of monomials that include variables
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.

13. Example 3A: Finding the GCF of Monomials
Find the GCF of each pair of monomials.
15x3 and 9x2
Write the prime factorization of each coefficient and write powers as products.
15x3 = 3  5  x  x  x
9x2 = 3  3  x  x
Align the common factors.
3  x  x = 3x2
Find the product of the common factors.
The GCF of 3x3 and 6x2 is 3x2.

14. Example 3B: Finding the GCF of Monomials
Find the GCF of each pair of monomials.
8x2 and 7y3
The GCF 8x2 and 7y is 1.

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

16. Check It Out! Example 3a
Find the GCF of each pair of monomials.
18g2 and 27g3
Write the prime factorization of each coefficient and write powers as products.
18g2 = 2  3  3  g  g
27g3 =  3  3  g  g  g
Align the common factors.
3  3  g  g
Find the product of the common factors.
The GCF of 18g2 and 27g3 is 9g2.

17. Find the GCF of each pair of monomials.
Check It Out! Example 3b
Find the GCF of each pair of monomials.
Write the prime factorization of each coefficient and write powers as products.
16a6 and 9b
16a6 = 2  2  2  2  a  a  a  a  a  a
9b =  3  b
Align the common factors.
The GCF of 16a6 and 7b is 1.
There are no common factors other than 1.

18. Check It Out! Example 4
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.

19. Check It Out! Example 4 Continued
Find the common factors of 36 and 48.
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
The GCF of 36 and 48 is 12.
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.

20. 36 pop CDs
12 CDs per shelf
= 3 shelves
48 country CDs
= 4 shelves
When the greatest possible number of CD types are on each shelf, there are 7 shelves in total.

21. 5 Minute Warm-Up
Write the prime factorization of each number.
Find the GCF of each pair monomial.
3. 18 and and 36
5. 12x and 28x x2 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?

22. Objective
Factor polynomials by using the greatest common factor.

23. Recall that the Distributive Property states that ab + ac =a(b + c)
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.

24. Example 1A: Factoring by Using the GCF
Factor each polynomial. Check your answer.
2x2 – 4
2x2 = 2  x  x
Find the GCF.
4 = 2  2 2
The GCF of 2x2 and 4 is 2.
Write terms as products using the GCF as a factor.
2x2 – (2  2)
2(x2 – 2)
Use the Distributive Property to factor out the GCF.
Multiply to check your answer.
Check 2(x2 – 2)
The product is the original polynomial. 
2x2 – 4

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

26. Example 1B: Factoring by Using the GCF
Factor each polynomial. Check your answer.
8x3 – 4x2 – 16x
8x3 = 2  2  2  x  x  x
Find the GCF.
4x2 = 2  2  x  x
16x = 2  2  2  2  x
The GCF of 8x3, 4x2, and 16x is 4x.
2  2  x = 4x
Write terms as products using the GCF as a factor.
2x2(4x) – x(4x) – 4(4x)
Use the Distributive Property to factor out the GCF.
4x(2x2 – x – 4)
Check
4x(2x2 – x – 4)
Multiply to check your answer.
The product is the original polynomials.
8x3 – 4x2 – 16x 

27. Example 1C: Factoring by Using the GCF
Factor each polynomial. Check your answer.
–14x – 12x2
– 1(14x + 12x2)
Both coefficients are negative. Factor out –1.
14x = 2   x
12x2 = 2  2  3  x  x
Find the GCF.
The GCF of 14x and 12x2 is 2x.
2  x = 2x
–1[7(2x) + 6x(2x)]
Write each term as a product using the GCF.
–1[2x(7 + 6x)]
Use the Distributive Property to factor out the GCF.
–2x(7 + 6x)

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

29. Example 1D: Factoring by Using the GCF
Factor each polynomial. Check your answer.
3x3 + 2x2 – 10
3x3 =  x  x  x
Find the GCF.
2x2 =  x  x
10 =  5
There are no common factors other than 1.
3x3 + 2x2 – 10
The polynomial cannot be factored further.

30.  Check It Out! Example 1a Factor each polynomial. Check your answer.
5b + 9b3
5b = 5  b
Find the GCF.
9b = 3  3  b  b  b
The GCF of 5b and 9b3 is b. b
Write terms as products using the GCF as a factor.
5(b) + 9b2(b)
Use the Distributive Property to factor out the GCF.
b(5 + 9b2)
Multiply to check your answer.
b(5 + 9b2)
Check
The product is the original polynomial.
5b + 9b3 

31. Check It Out! Example 1b
Factor each polynomial. Check your answer.
9d2 – 82
9d2 = 3  3  d  d
Find the GCF.
82 =  2  2  2  2  2
There are no common factors other than 1.
9d2 – 82
The polynomial cannot be factored further.

32. Check It Out! Example 1c
Factor each polynomial. Check your answer.
–18y3 – 7y2
– 1(18y3 + 7y2)
Both coefficients are negative. Factor out –1.
18y3 = 2  3  3  y  y  y
Find the GCF.
7y2 = 7  y  y
y  y = y2
The GCF of 18y3 and 7y2 is y2.
Write each term as a product using the GCF.
–1[18y(y2) + 7(y2)]
–1[y2(18y + 7)]
Use the Distributive Property to factor out the GCF..
–y2(18y + 7)

33. Factor each polynomial. Check your answer.
Check It Out! Example 1d
Factor each polynomial. Check your answer.
8x4 + 4x3 – 2x2
8x4 = 2  2  2  x  x  x  x
4x3 = 2  2  x  x  x
Find the GCF.
2x2 = 2  x  x
2  x  x = 2x2
The GCF of 8x4, 4x3 and –2x2 is 2x2.
Write terms as products using the GCF as a factor.
4x2(2x2) + 2x(2x2) –1(2x2)
2x2(4x2 + 2x – 1)
Use the Distributive Property to factor out the GCF.
Check 2x2(4x2 + 2x – 1)
Multiply to check your answer.
8x4 + 4x3 – 2x2
The product is the original polynomial.

34. 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.

35. 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
The GCF of 9x2 and 6x is 3x.
= 3x(3x) + 2(3x)
Write each term as a product using the GCF as a factor.
= 3x(3x + 2)
Use the Distributive Property to factor out the GCF.
Possible expressions for the dimensions of the squash court are 3x m and (3x + 2) m.

36. Sometimes the GCF of terms is a binomial
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.

37. Example 3: Factoring Out a Common Binomial Factor
Factor each expression.
A. 5(x + 2) + 3x(x + 2)
The terms have a common binomial factor of (x + 2).
5(x + 2) + 3x(x + 2)
(x + 2)(5 + 3x)
Factor out (x + 2).
B. –2b(b2 + 1)+ (b2 + 1)
The terms have a common binomial factor of (b2 + 1).
–2b(b2 + 1) + (b2 + 1)
–2b(b2 + 1) + 1(b2 + 1)
(b2 + 1) = 1(b2 + 1)
(b2 + 1)(–2b + 1)
Factor out (b2 + 1).

38. Check It Out! Example 3
Factor each expression.
a. 4s(s + 6) – 5(s + 6)
The terms have a common binomial factor of (s + 6).
4s(s + 6) – 5(s + 6)
(4s – 5)(s + 6)
Factor out (s + 6).
b. 7x(2x + 3) + (2x + 3)
7x(2x + 3) + (2x + 3)
The terms have a common binomial factor of (2x + 3).
7x(2x + 3) + 1(2x + 3)
(2x + 1) = 1(2x + 1)
(2x + 3)(7x + 1)
Factor out (2x + 3).

39. You may be able to factor a polynomial by grouping
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.

40. Example 4A: Factoring by Grouping
Factor each polynomial by grouping. Check your answer.
6h4 – 4h3 + 12h – 8
Group terms that have a common number or variable as a factor.
(6h4 – 4h3) + (12h – 8)
2h3(3h – 2) + 4(3h – 2)
Factor out the GCF of each group.
2h3(3h – 2) + 4(3h – 2)
(3h – 2) is another common factor.
(3h – 2)(2h3 + 4)
Factor out (3h – 2).

41. Example 4A Continued
Factor each polynomial by grouping. Check your answer.
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 
The product is the original polynomial.

42. Example 4B: Factoring by Grouping
Factor each polynomial by grouping. Check your answer.
5y4 – 15y3 + y2 – 3y
(5y4 – 15y3) + (y2 – 3y)
Group terms.
Factor out the GCF of each group.
5y3(y – 3) + y(y – 3)
5y3(y – 3) + y(y – 3)
(y – 3) is a common factor.
(y – 3)(5y3 + y)
Factor out (y – 3).

43. Check It Out! Example 4a
Factor each polynomial by grouping. Check your answer.
6b3 + 8b2 + 9b + 12
(6b3 + 8b2) + (9b + 12)
Group terms.
2b2(3b + 4) + 3(3b + 4)
Factor out the GCF of each group.
(3b + 4) is a common factor.
2b2(3b + 4) + 3(3b + 4)
(3b + 4)(2b2 + 3)
Factor out (3b + 4).

44. Check It Out! Example 4b
Factor each polynomial by grouping. Check your answer.
4r3 + 24r + r2 + 6
(4r3 + 24r) + (r2 + 6)
Group terms.
4r(r2 + 6) + 1(r2 + 6)
Factor out the GCF of each group.
4r(r2 + 6) + 1(r2 + 6)
(r2 + 6) is a common factor.
(r2 + 6)(4r + 1)
Factor out (r2 + 6).

45. If two quantities are opposites, their sum is 0.
(5 – x) + (x – 5)
5 – x + x – 5
– x + x + 5 – 5
0 + 0
Helpful Hint

46. Recognizing opposite binomials can help you factor polynomials
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)
Distributive Property.
= –x + 5
Simplify.
= 5 – x
Commutative Property of Addition.
So, (5 – x) = –1(x – 5)

47. Example 5: Factoring with Opposites
Factor 2x3 – 12x – 3x
2x3 – 12x – 3x
(2x3 – 12x2) + (18 – 3x)
Group terms.
2x2(x – 6) + 3(6 – x)
Factor out the GCF of each group.
2x2(x – 6) + 3(–1)(x – 6)
Write (6 – x) as –1(x – 6).
2x2(x – 6) – 3(x – 6)
Simplify. (x – 6) is a common factor.
(x – 6)(2x2 – 3)
Factor out (x – 6).

48. Check It Out! Example 5a
Factor each polynomial. Check your answer.
15x2 – 10x3 + 8x – 12
(15x2 – 10x3) + (8x – 12)
Group terms.
Factor out the GCF of each group.
5x2(3 – 2x) + 4(2x – 3)
5x2(3 – 2x) + 4(–1)(3 – 2x)
Write (2x – 3) as –1(3 – 2x).
Simplify. (3 – 2x) is a common factor.
5x2(3 – 2x) – 4(3 – 2x)
(3 – 2x)(5x2 – 4)
Factor out (3 – 2x).

49. 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 – 2p p – 18

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

51. 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.

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

53. Example 1A: Factoring Trinomials by Guess and Check
Factor x2 + 15x + 36 by guess and check.
( )( )
Write two sets of parentheses.
(x + )(x + )
The first term is x2, so the variable terms have a coefficient of 1.
The constant term in the trinomial is 36. 
Try factors of 36 for the constant terms in the binomials.
(x + 1)(x + 36) = x2 + 37x + 36 
(x + 2)(x + 18) = x2 + 20x + 36 
(x + 3)(x + 12) = x2 + 15x + 36
The factors of x2 + 15x + 36 are (x + 3)(x + 12).
x2 + 15x + 36 = (x + 3)(x + 12)

54. When you multiply two binomials, multiply:
First terms
Outer terms
Inner terms
Last terms
Remember!

55.     Check It Out! Example 1a
Factor each trinomial by guess and check.
x2 + 10x + 24
( )( )
Write two sets of parentheses.
(x + )(x + )
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
Try factors of 24 for the constant terms in the binomials. 
(x + 2)(x + 12) = x2 + 14x + 24 
(x + 3)(x + 8) = x2 + 11x + 24 
(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)

56. 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). x2 12
(x + 3)(x +4) = x2 + 7x + 12 3x 4x
The coefficient of the middle term is the sum of 3 and 4. The third term is the product of 3 and 4.

58. When c is positive, its factors have the same sign
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.

59. Example 2A: Factoring x2 + bx + c When c is Positive
Factor each trinomial. Check your answer.
x2 + 6x + 5
(x + )(x + )
b = 6 and c = 5; look for factors of 5 whose sum is 6.
Factors of 5 Sum
1 and 
The factors needed are 1 and 5.
(x + 1)(x + 5)
Check (x + 1)(x + 5) = x2 + x + 5x + 5
Use the FOIL method.
= x2 + 6x + 5 
The product is the original polynomial.

60. Example 2C: Factoring x2 + bx + c When c is Positive
Factor each trinomial. Check your answer.
x2 – 8x + 15
(x + )(x + )
b = –8 and c = 15; look for factors of 15 whose sum is –8.
Factors of –15 Sum
–1 and –15 –16 
–3 and –5 –8 
The factors needed are –3 and –5 .
(x – 3)(x – 5)
Check (x – 3)(x – 5 ) = x2 – 3x – 5x + 15
Use the FOIL method.
= x2 – 8x + 15 
The product is the original polynomial.

61.    Check It Out! Example 2b
Factor each trinomial. Check your answer.
x2 – 5x + 6
(x + )(x+ )
b = –5 and c = 6; look for factors of 6 whose sum is –5.
Factors of 6 Sum
–1 and –6 –7 
–2 and –3 –5 
The factors needed are –2 and –3.
(x – 2)(x – 3)
Check (x – 2)(x – 3) = x2 –2x – 3x + 6
Use the FOIL method.
= x2 – 5x + 6 
The product is the original polynomial.

62.    Check It Out! Example 2d
Factor each trinomial. Check your answer.
x2 – 13x + 40
b = –13 and c = 40; look for factors of 40 whose sum is –13.
(x + )(x+ )
Factors of 40 Sum 
–2 and –20 –22
–4 and –10 –14 
–5 and – –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.
= x2 – 13x + 40 
The product is the original polynomial.

63. When c is negative, its factors have opposite signs
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.

64. Example 3A: Factoring x2 + bx + c When c is Negative
Factor each trinomial.
x2 + x – 20
b = 1 and c = –20; look for factors of –20 whose sum is 1. The factor with the greater absolute value is positive.
(x + )(x + )
Factors of –20 Sum 
–1 and
–2 and 
–4 and
The factors needed are +5 and –4.
(x – 4)(x + 5)

65. Example 3B: Factoring x2 + bx + c When c is Negative
Factor each trinomial.
x2 – 3x – 18
b = –3 and c = –18; look for factors of –18 whose sum is –3. The factor with the greater absolute value is negative.
(x + )(x + )
Factors of –18 Sum 
1 and – –17
2 and – – 7 
3 and – – 3
The factors needed are 3 and –6.
(x – 6)(x + 3)

66. 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.

67. Example 4A: Evaluating Polynomials
Factor y2 + 10y 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 Sum
1 and 
3 and 
The factors needed are 3 and 7.
(y + 3)(y + 7)

68. 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 1 2 3 4
y2 + 10y + 21
(0) + 21 = 21 y 1 2 3 4
(1) + 21 = 32
(2) + 21 = 45
(3) + 21 = 60
(4) + 21 = 77
The original polynomial and the factored form have the same value for the given values of n.

69. Lesson Quiz: Part I
Factor each trinomial.
1. x2 – 11x + 30
2. x2 + 10x + 9
3. x2 – 6x – 27
4. x2 + 14x – 32
(x – 5)(x – 6)
(x + 1)(x + 9)
(x – 9)(x + 3)
(x + 16)(x – 2)

70. 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 – 32
5. z2 + 15z + 36
6. h2 – 17h + 72
2x2 + 3x – 14
6y2 + 35y + 36
3n2 – 26n + 35
(x – 4)(x + 8)
(z + 3)(z + 12)
(h – 8)(h – 9)

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

72. In the previous lesson you factored trinomials of the form x2 + bx + c
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.

73. 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

74. 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

75. Example 1: Factoring ax2 + bx + c by Guess and Check
Factor 6x2 + 11x + 4 by guess and check.
( )( )
Write two sets of parentheses.
( x + )( x + )
The first term is 6x2, so at least one variable term has a coefficient other than 1.
The coefficient of the x2 term is 6. The constant term in the trinomial is 4.
(2x + 4)(3x + 1) = 6x2 + 14x + 4 
(1x + 4)(6x + 1) = 6x2 + 25x + 4 
Try factors of 6 for the coefficients and factors of 4 for the constant terms.
(1x + 2)(6x + 2) = 6x2 + 14x + 4 
(1x + 1)(6x + 4) = 6x2 + 10x + 4 
(3x + 4)(2x + 1) = 6x2 + 11x + 4 

76. 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.
Product = a
Product = c
Sum of outer and inner products = b
( X + )( x + ) = ax2 + bx + c

77. 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.
Product = a
Product = c
Sum of outer and inner products = b
( X + )( x + ) = ax2 + bx + c

78. Example 2A: Factoring ax2 + bx + c When c is Positive
Factor each trinomial. Check your answer.
2x2 + 17x + 21
a = 2 and c = 21, Outer + Inner = 17.
( x + )( x + )
Factors of 2 Factors of 21 Outer + Inner
1 and 2
1 and 21
1(21) + 2(1) = 23
21 and 1
1(1) + 2(21) = 43
3 and 7
1(7) + 2(3) = 13
7 and 3
1(3) + 2(7) = 17  
(x + 7)(2x + 3)
Use the Foil method.
Check (x + 7)(2x + 3) = 2x2 + 3x + 14x + 21
= 2x2 + 17x + 21 

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

80. Example 2B: Factoring ax2 + bx + c When c is Positive
Factor each trinomial. Check your answer.
3x2 – 16x + 16
a = 3 and c = 16,
Outer + Inner = –16 .
( x + )( x + )
Factors of 3 Factors of 16 Outer + Inner
1 and 3
–1 and –16
1(–16) + 3(–1) = –19
– 2 and – 8
1( – 8) + 3(–2) = –14
– 4 and – 4
1( – 4) + 3(– 4)= –16  
(x – 4)(3x – 4)
Use the Foil method.
Check (x – 4)(3x – 4) = 3x2 – 4x – 12x + 16
= 3x2 – 16x + 16 

81.    Check It Out! Example 2b
Factor each trinomial. Check your answer.
9x2 – 15x + 4
a = 9 and c = 4,
Outer + Inner = –15.
( x + )( x + )
Factors of 9 Factors of 4 Outer + Inner
3 and 3
–1 and – 4
3(–4) + 3(–1) = –15
– 2 and – 2
3(–2) + 3(–2) = –12
– 4 and – 1
3(–1) + 3(– 4)= –15  
(3x – 4)(3x – 1)
Use the Foil method.
Check (3x – 4)(3x – 1) = 9x2 – 3x – 12x + 4
= 9x2 – 15x + 4 

82.    Check It Out! Example 2c
Factor each trinomial. Check your answer.
3x2 + 13x + 12
a = 3 and c = 12,
Outer + Inner = 13.
( x + )( x + )
Factors of 3 Factors of 12 Outer + Inner
1 and 3
1 and 12
1(12) + 3(1) = 15
2 and 6
1(6) + 3(2) = 12
3 and 4
1(4) + 3(3) = 13  
(x + 3)(3x + 4)
Use the Foil method.
Check (x + 3)(3x + 4) = 3x2 + 4x + 9x + 12
= 3x2 + 13x + 12 

83. 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.

84. Example 3A: Factoring ax2 + bx + c When c is Negative
Factor each trinomial. Check your answer.
3n2 + 11n – 4
a = 3 and c = – 4,
Outer + Inner = 11 .
( y + )( y+ )
Factors of 3 Factors of 4 Outer + Inner
1 and 3
–1 and 4
1(4) + 3(–1) = 1
–2 and 2
1(2) + 3(–2) = – 4
–4 and 1
1(1) + 3(–4) = –11 
4 and –1
1(–1) + 3(4) = 11 
(n + 4)(3n – 1)
Use the Foil method.
Check (n + 4)(3n – 1) = 3n2 – n + 12n – 4
= 3n2 + 11n – 4 

85. Example 3B: Factoring ax2 + bx + c When c is Negative
Factor each trinomial. Check your answer.
2x2 + 9x – 18
a = 2 and c = –18,
Outer + Inner = 9 .
( x + )( x+ )
Factors of 2 Factors of – 18 Outer + Inner
1 and 2
18 and –1
1(– 1) + 2(18) = 35
9 and –2
1(– 2) + 2(9) = 16
6 and –3
1(– 3) + 2(6) = 9  
(x + 6)(2x – 3)
Use the Foil method.
Check (x + 6)(2x – 3) = 2x2 – 3x + 12x – 18
= 2x2 + 9x – 18 

86. Example 3C: Factoring ax2 + bx + c When c is Negative
Factor each trinomial. Check your answer.
4x2 – 15x – 4
a = 4 and c = –4,
Outer + Inner = –15.
( x + )( x+ )
Factors of 4 Factors of – 4 Outer + Inner
1 and 4
–1 and 4
1(4) – 1(4) = 0
–2 and 2
1(2) – 2(4) = –6
–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 

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

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

89. Example 4A: Factoring ax2 + bx + c When a is Negative
Factor –2x2 – 5x – 3.
–1(2x2 + 5x + 3)
Factor out –1.
a = 2 and c = 3;
Outer + Inner = 5
–1( x + )( x+ )
Factors of 2 Factors of Outer + Inner
1 and 2
3 and 1
1(1) + 3(2) = 7  
1 and 3
1(3) + 1(2) = 5
(x + 1)(2x + 3)
–1(x + 1)(2x + 3)

90.   Check It Out! Example 4a Factor each trinomial. Check your answer.
–6x2 – 17x – 12
Factor out –1.
–1(6x2 + 17x + 12)
a = 6 and c = 12;
Outer + Inner = 17
–1( x + )( x+ )
Factors of 6 Factors of Outer + Inner
2 and 3
4 and 3
2(3) + 3(4) = 18  
3 and 4
2(4) + 3(3) = 17
(2x + 3)(3x + 4)
–1(2x + 3)(3x + 4)

91.   Check It Out! Example 4b Factor each trinomial. Check your answer.
–3x2 – 17x – 10
Factor out –1.
–1(3x2 + 17x + 10)
a = 3 and c = 10;
Outer + Inner = 17)
–1( x + )( x+ )
Factors of 3 Factors of Outer + Inner
1 and 3
2 and 5
1(5) + 3(2) = 11  
5 and 2
1(2) + 3(5) = 17
(3x + 2)(x + 5)
–1(3x + 2)(x + 5)

92. 5 Minute Warm-Up
Factor each trinomial.
1. 5x2 + 17x + 6
2. 2x2 + 5x – 12
3. 6x2 – 23x + 7
4. –4x2 + 11x + 20
5. –2x2 + 7x – 3
6. 8x2 + 27x + 9
(5x + 2)(x + 3)
(2x– 3)(x + 4)
(3x – 1)(2x – 7)
(–x + 4)(4x + 5)
(–2x + 1)(x – 3)
(8x + 3)(x + 3)

93. 5 Minute Warm-Up
Factor each expression.
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

94. Objectives Factor perfect-square trinomials.
Factor the difference of two squares.

95. 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.
9x x
3x 3x
2(3x 2)
2 2 •

97. 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
8 8
3x 3x
2(3x 8) 
2(3x 8) ≠ –15x. 
9x2 – 15x + 64 is not a perfect-square trinomial because –15x ≠ 2(3x  8).

98. Example 1B: Recognizing and Factoring Perfect-Square Trinomials
Determine whether each trinomial is a perfect square. If so, factor. If not explain.
81x2 + 90x + 25
81x2 + 90x + 25
5 5
9x 9x
2(9x 5) ●
The trinomial is a perfect square. Factor.

99. Example 1B Continued
Determine whether each trinomial is a perfect square. If so, factor. If not explain.
Method 2 Use the rule.
81x x + 25
a = 9x, b = 5
(9x)2 + 2(9x)(5) + 52
Write the trinomial as a2 + 2ab + b2.
Write the trinomial as (a + b)2.
(9x + 5)2

100. Example 1C: Recognizing and Factoring Perfect-Square Trinomials
Determine whether each trinomial is a perfect square. If so, factor. If not explain.
36x2 – 10x + 14
36x2 – 10x + 14
The trinomial is not a perfect-square because 14 is not a perfect square.
36x2 – 10x + 14 is not a perfect-square trinomial.

101. Example 2: Problem-Solving Application
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.

102. Example 2 Continued 2
Make a Plan
The formula for the area of a square is
area = (side)2.
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.

103. Example 2 Continued
Solve 3
a = 4x, b = 3
16x2 – 24x + 9
Write the trinomial as
a2 – 2ab + b2.
(4x)2 – 2(4x)(3) + 32
(4x – 3)2
Write the trinomial as (a + b)2.
16x2 – 24x + 9 = (4x – 3)(4x – 3)
The side length of the tablecloth is (4x – 3) in. and (4x – 3) in.

104. Example 2 Continued
Write a formula for the perimeter of the tablecloth.
Write the formula for the perimeter of a square.
P = 4s
= 4(4x – 3)
Substitute the side length for s.
= 16x – 12
Distribute 4.
An expression for the perimeter of the tablecloth in inches is 16x – 12.

105. Example 2 Continued
Evaluate the expression when x = 11.
P = 16x – 12
= 16(11) – 12
Substitute 11 for x.
= 164
When x = 11 in. the perimeter of the tablecloth is 164 in.

106. A polynomial is a difference of two squares if:
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 

108. 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

109. 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.

110. Example 3B: Recognizing and Factoring the Difference of Two Squares
Determine whether each binomial is a difference of two squares. If so, factor. If not, explain.
100x2 – 4y2
100x2 – 4y2
2y 2y 
10x 10x
The polynomial is a difference of two squares.
(10x)2 – (2y)2
a = 10x, b = 2y
(10x + 2y)(10x – 2y)
Write the polynomial as (a + b)(a – b).
100x2 – 4y2 = (10x + 2y)(10x – 2y)

111. Example 3C: Recognizing and Factoring the Difference of Two Squares
Determine whether each binomial is a difference of two squares. If so, factor. If not, explain.
x4 – 25y6
x4 – 25y6
5y3 5y3 
x2 x2
The polynomial is a difference of two squares.
(x2)2 – (5y3)2
a = x2, b = 5y3
Write the polynomial as (a + b)(a – b).
(x2 + 5y3)(x2 – 5y3)
x4 – 25y6 = (x2 + 5y3)(x2 – 5y3)

112. The polynomial is a difference of two squares.
Check It Out! Example 3a
Determine whether each binomial is a difference of two squares. If so, factor. If not, explain.
1 – 4x2
1 – 4x2
2x 2x 
The polynomial is a difference of two squares.
(1)2 – (2x)2
a = 1, b = 2x
(1 + 2x)(1 – 2x)
Write the polynomial as (a + b)(a – b).
1 – 4x2 = (1 + 2x)(1 – 2x)

113. The polynomial is a difference of two squares.
Check It Out! Example 3b
Determine whether each binomial is a difference of two squares. If so, factor. If not, explain.
p8 – 49q6
p8 – 49q6
7q3 7q3 ●
p4 p4
The polynomial is a difference of two squares.
(p4)2 – (7q3)2
a = p4, b = 7q3
(p4 + 7q3)(p4 – 7q3)
Write the polynomial as
(a + b)(a – b).
p8 – 49q6 = (p4 + 7q3)(p4 – 7q3)

114. Check It Out! Example 3c
Determine whether each binomial is a difference of two squares. If so, factor. If not, explain.
16x2 – 4y5
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.

115. 5 Minute Warm-up
Determine whether each trinomial is a perfect square. If so factor. If not, explain.
64x2 – 40x + 25
2. 121x2 – 44x + 4
3. 49x x + 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.
Not a perfect-square trinomial because –40x ≠ 2(8x  5).
(11x – 2)2
(7x2 + 10)2
P = 28x + 16; 156 ft

116. Objectives Choose an appropriate method for factoring a polynomial.
Combine methods for factoring a polynomial.

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

118. 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)
6x – 4 can be further factored.
6x2(3x – 2)
Factor out 2, the GCF of 6x and – 4.
6x2(3x – 2) is completely factored.
B. (x2 + 1)(x – 5)
(x2 + 1)(x – 5)
Neither x2 +1 nor x – 5 can be factored further.
(x2 + 1)(x – 5) is completely factored.

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

120. Check It Out! Example 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.
5x2(x – 1) is completely factored.
B. (4x + 4)(x + 1)
(4x + 4)(x + 1)
4x + 4 can be further factored.
4(x + 1)(x + 1)
Factor out 4, the GCF of 4x and 4.
4(x + 1)2 is completely factored.

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

122. Example 2A: Factoring by GCF and Recognizing Patterns
Factor 10x2 + 48x + 32 completely. Check your answer.
10x2 + 48x + 32
2(5x2 + 24x + 16)
Factor out the GCF.
2(5x + 4)(x + 4)
Factor remaining trinomial.
Check 2(5x + 4)(x + 4) = 2(5x2 + 20x + 4x + 16)
= 10x2 + 40x + 8x + 32
= 10x2 + 48x + 32 

123. Example 2B: Factoring by GCF and Recognizing Patterns
Factor 8x6y2 – 18x2y2 completely. Check your answer.
8x6y2 – 18x2y2
Factor out the GCF. 4x4 – 9 is a perfect-square trinomial of the form a2 – b2.
2x2y2(4x4 – 9)
2x2y2(2x2 – 3)(2x2 + 3)
a = 2x, b = 3
Check 2x2y2(2x2 – 3)(2x2 + 3) = 2x2y2(4x4 – 9)
= 8x6y2 – 18x2y2 

124. Check It Out! Example 2a
Factor each polynomial completely. Check your answer.
4x3 + 16x2 + 16x
Factor out the GCF. x2 + 4x + 4 is a perfect-square trinomial of the form a2 + 2ab + b2.
4x3 + 16x2 + 16x
4x(x2 + 4x + 4)
4x(x + 2)2
a = x, b = 2
Check 4x(x + 2)2 = 4x(x2 + 2x + 2x + 4)
= 4x(x2 + 4x + 4)
= 4x3 + 16x2 + 16x 

125. Check It Out! Example 2b
Factor each polynomial completely. Check your answer.
2x2y – 2y3
Factor out the GCF. 2y(x2 – y2) is a perfect-square trinomial of the form a2 – b2.
2x2y – 2y3
2y(x2 – y2)
2y(x + y)(x – y)
a = x, b = y
Check 2y(x + y)(x – y) = 2y(x2 + xy – xy – y2)
= 2x2y +2xy2 – 2xy2 – 2y3
= 2x2y – 2y3

126. 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

127. Example 3A: Factoring by Multiple Methods
Factor each polynomial completely.
9x2 + 3x – 2
The GCF is 1 and there is no pattern.
9x2 + 3x – 2
( x + )( x + )
a = 9 and c = –2;
Outer + Inner = 3 
Factors of 9 Factors of 2 Outer + Inner
1 and 9
1 and –2
1(–2) + 1(9) = 7
3 and 3
3(–2) + 1(3) = –3
–1 and 2
3(2) + 3(–1) = 3 
(3x – 1)(3x + 2)

128. Example 3B: Factoring by Multiple Methods
Factor each polynomial completely.
12b3 + 48b2 + 48b
The GCF is 12b; (b2 + 4b + 4) is a perfect-square trinomial in the form of
a2 + 2ab + b2.
12b(b2 + 4b + 4)
(x + )(x + )
Factors of 4 Sum
1 and
2 and  
a = 2 and c = 2
12b(b + 2)(b + 2)
12b(b + 2)2

129. Example 3C: Factoring by Multiple Methods
Factor each polynomial completely.
4y2 + 12y – 72
Factor out the GCF. There is no pattern. b = 3 and c = –18; look for factors of –18 whose sum is 3.
4(y2 + 3y – 18)
(y + )(y + ) 
Factors of –18 Sum
–1 and
–2 and 
–3 and
The factors needed are –3 and 6
4(y – 3)(y + 6)

130. Example 3D: Factoring by Multiple Methods.
Factor each polynomial completely.
(x4 – x2)
x2(x2 – 1)
Factor out the GCF.
x2(x + 1)(x – 1)
x2 – 1 is a difference of two squares.

131.  Check It Out! Example 3b Factor each polynomial completely.
2p5 + 10p4 – 12p3
Factor out the GCF. There is no pattern. b = 5 and c = –6; look for factors of –6 whose sum is 5.
2p3(p2 + 5p – 6)
(p + )(p + ) 
Factors of – 6 Sum
– 1 and
The factors needed are –1 and 6
2p3(p + 6)(p – 1)

132.   Check It Out! Example 3c Factor each polynomial completely.
9q6 + 30q5 + 24q4
Factor out the GCF. There is no pattern.
3q4(3q2 + 10q + 8)
( q + )( q + )
a = 3 and c = 8;
Outer + Inner = 10 
Factors of 3 Factors of 8 Outer + Inner
3 and 1
1 and 8
3(8) + 1(1) = 25
2 and 4
3(4) + 1(2) = 14
4 and 2
3(2) + 1(4) = 10 
3q4(3q + 4)(q + 2)

134. Lesson Quiz
Tell whether the polynomial is completely factored. If not, factor it.
1. (x + 3)(5x + 10) x2(x2 + 9)
no; 5(x+ 3)(x + 2)
completely factored
Factor each polynomial completely. Check your answer.
3. x3 + 4x2 + 3x x2 + 16x – 48
4(x + 6)(x – 2)
(x + 4)(x2 + 3)
5. 18x2 – 3x – 3
6. 18x2 – 50y2
3(3x + 1)(2x – 1)
2(3x + 5y)(3x – 5y)
7. 5x – 20x3 + 7 – 28x2
(1 + 2x)(1 – 2x)(5x + 7)