1. Factoring Polynomials
3-4
Factoring Polynomials
Warm Up
Lesson Presentation
Lesson Quiz
Holt McDougal Algebra 2
Holt Algebra 2

2. Warm Up Factor each expression. 1. 3x – 6y 3(x – 2y) 2. a2 – b2
(a + b)(a – b)

3. Objectives
Use the Factor Theorem to determine factors of a polynomial.
Factor the sum and difference of two cubes.

4. Recall that if a number is divided by any of its factors, the remainder is 0. Likewise, if a polynomial is divided by any of its factors, the remainder is 0.
The Remainder Theorem states that if a polynomial is divided by (x – a), the remainder is the value of the function at a. So, if (x – a) is a factor of P(x), then P(a) = 0.

5. Example 1: Determining Whether a Linear Binomial is a Factor
Determine whether the given binomial is a factor of the polynomial P(x).
A. (x + 1); (x2 – 3x + 1)
B. (x + 2); (3x4 + 6x3 – 5x – 10)
Find P(–1) by synthetic substitution.
Find P(–2) by synthetic substitution. –1
1 –3 1 –1 4 –2
–5 –10 1 –4 5 –6 10 3 –5
P(–1) = 5
P(–1) ≠ 0, so (x + 1) is not a factor of P(x) = x2 – 3x + 1.
P(–2) = 0, so (x + 2) is a factor of P(x) = 3x4 + 6x3 – 5x – 10.

6. Find P(–2) by synthetic substitution.
Check It Out! Example 1
Determine whether the given binomial is a factor of the polynomial P(x).
a. (x + 2); (4x2 – 2x + 5)
b. (3x – 6); (3x4 – 6x3 + 6x2 + 3x – 30)
Find P(–2) by synthetic substitution.
Divide the polynomial by 3, then find P(2) by synthetic substitution. –2
4 –2 5 –8 20 2
1 – –10 4
–10 25 2 4 10
P(–2) = 25 1 2 5
P(–2) ≠ 0, so (x + 2) is not a factor of P(x) = 4x2 – 2x + 5.
P(2) = 0, so (3x – 6) is a factor of P(x) = 3x4 – 6x3 + 6x2 + 3x – 30.

7. You are already familiar with methods for factoring quadratic expressions. You can factor polynomials of higher degrees using many of the same methods you learned in Lesson 5-3.

8. Example 2: Factoring by Grouping
Factor: x3 – x2 – 25x + 25.
(x3 – x2) + (–25x + 25)
Group terms.
Factor common monomials from each group.
x2(x – 1) – 25(x – 1)
Factor out the common binomial (x – 1).
(x – 1)(x2 – 25)
Factor the difference of squares.
(x – 1)(x – 5)(x + 5)

9. Check It Out! Example 2a
Factor: x3 – 2x2 – 9x + 18.
(x3 – 2x2) + (–9x + 18)
Group terms.
Factor common monomials from each group.
x2(x – 2) – 9(x – 2)
Factor out the common binomial (x – 2).
(x – 2)(x2 – 9)
Factor the difference of squares.
(x – 2)(x – 3)(x + 3)

10. Check It Out! Example 2b
Factor: 2x3 + x2 + 8x + 4.
(2x3 + x2) + (8x + 4)
Group terms.
Factor common monomials from each group.
x2(2x + 1) + 4(2x + 1)
Factor out the common binomial (2x + 1).
(2x + 1)(x2 + 4)
(2x + 1)(x2 + 4)

11. Just as there is a special rule for factoring the difference of two squares, there are special rules for factoring the sum or difference of two cubes.

12. Example 3A: Factoring the Sum or Difference of Two Cubes
Factor the expression.
4x x
4x(x3 + 27)
Factor out the GCF, 4x.
4x(x3 + 33)
Rewrite as the sum of cubes.
Use the rule a3 + b3 = (a + b)  (a2 – ab + b2).
4x(x + 3)(x2 – x  )
4x(x + 3)(x2 – 3x + 9)

13. Example 3B: Factoring the Sum or Difference of Two Cubes
Factor the expression.
125d3 – 8
Rewrite as the difference of cubes.
(5d)3 – 23
(5d – 2)[(5d)2 + 5d  ]
Use the rule a3 – b3 = (a – b)  (a2 + ab + b2).
(5d – 2)(25d2 + 10d + 4)

14. Check It Out! Example 3a
Factor the expression.
8 + z6
Rewrite as the difference of cubes.
(2)3 + (z2)3
(2 + z2)[(2)2 – 2  z + (z2)2]
Use the rule a3 + b3 = (a + b)  (a2 – ab + b2).
(2 + z2)(4 – 2z + z4)

15. Check It Out! Example 3b
Factor the expression.
2x5 – 16x2
2x2(x3 – 8)
Factor out the GCF, 2x2.
Rewrite as the difference of cubes.
2x2(x3 – 23)
Use the rule a3 – b3 = (a – b)  (a2 + ab + b2).
2x2(x – 2)(x2 + x  )
2x2(x – 2)(x2 + 2x + 4)

16. Lesson Quiz
1. x – 1; P(x) = 3x2 – 2x + 5
P(1) ≠ 0, so x – 1 is not a factor of P(x).
2. x + 2; P(x) = x3 + 2x2 – x – 2
P(2) = 0, so x + 2 is a factor of P(x).
3. x3 + 3x2 – 9x – 27
(x + 3)(x + 3)(x – 3)
4. x3 + 3x2 – 28x – 60
(x + 6)(x – 5)(x + 2)
4. 64p3 – 8q3
8(2p – q)(4p2 + 2pq + q2)