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Holt Algebra 2
6-4 Factoring Polynomials 6-4 Factoring Polynomials
Holt Algebra 2
Warm Up
Lesson Presentation
Lesson Quiz
Holt Algebra 2
6-4 Factoring Polynomials
Warm Up Factor each expression.
1. 3x – 6y
2. a2 – b2
3. (x – 1)(x + 3)
4. (a + 1)(a2 + 1)
x2 + 2x – 3
3(x – 2y)
(a + b)(a – b)
a3 + a2 + a + 1
Find each product.
Holt Algebra 2
6-4 Factoring Polynomials
Use the Factor Theorem to determine factors of a polynomial.
Factor the sum and difference of two cubes.
Objectives
Holt Algebra 2
6-4 Factoring Polynomials
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.
Holt Algebra 2
6-4 Factoring Polynomials
Determine whether the given binomial is a factor of the polynomial P(x).
Example 1: Determining Whether a Linear Binomial is
a Factor
A. (x + 1); (x2 – 3x + 1)
Find P(–1) by synthetic substitution.
1 –3 1 –1
–1
1 5 –4
4
P(–1) = 5
P(–1) ≠ 0, so (x + 1) is not a factor of P(x) = x2 – 3x + 1.
B. (x + 2); (3x4 + 6x3 – 5x – 10)
Find P(–2) by synthetic substitution.
3 6 0 –5 –10 –2
–6
3 0
10 0 0
–5 0 0
P(–2) = 0, so (x + 2) is a factor of P(x) = 3x4 + 6x3 – 5x – 10.
Holt Algebra 2
6-4 Factoring Polynomials
Check It Out! Example 1
Determine whether the given binomial is a factor of the polynomial P(x).
a. (x + 2); (4x2 – 2x + 5)
Find P(–2) by synthetic substitution.
4 –2 5 –2
–8
4 25 –10
20
P(–2) = 25
P(–2) ≠ 0, so (x + 2) is not a factor of P(x) = 4x2 – 2x + 5.
b. (3x – 6); (3x4 – 6x3 + 6x2 + 3x – 30)
1 –2 2 1 –10 2
2
1 0
10 4 0
5 2 0
P(2) = 0, so (3x – 6) is a
factor of P(x) = 3x4 – 6x3 +
6x2 + 3x – 30.
Divide the polynomial by 3,
then find P(2) by synthetic
substitution.
Holt Algebra 2
6-4 Factoring Polynomials
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.
Holt Algebra 2
6-4 Factoring Polynomials
Factor: x3 – x2 – 25x + 25.
Example 2: Factoring by Grouping
Group terms. (x3 – x2) + (–25x + 25)
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)
Holt Algebra 2
6-4 Factoring Polynomials
Example 2 Continued
Check Use the table feature of your calculator to compare the original expression and the factored form.
The table shows that the original function and the factored form have the same function values.
Holt Algebra 2
6-4 Factoring Polynomials
Check It Out! Example 2a
Factor: x3 – 2x2 – 9x + 18.
Group terms. (x3 – 2x2) + (–9x + 18)
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)
Holt Algebra 2
6-4 Factoring Polynomials
Check It Out! Example 2a Continued
Check Use the table feature of your calculator to compare the original expression and the factored form.
The table shows that the original function and the factored form have the same function values.
Holt Algebra 2
6-4 Factoring Polynomials
Check It Out! Example 2b
Factor: 2x3 + x2 + 8x + 4.
Group terms. (2x3 + x2) + (8x + 4)
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)
Holt Algebra 2
6-4 Factoring Polynomials
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.
Holt Algebra 2
6-4 Factoring Polynomials
Example 3A: Factoring the Sum or Difference of Two
Cubes
Factor the expression.
4x4 + 108x
Factor out the GCF, 4x. 4x(x3 + 27)
Rewrite as the sum of cubes. 4x(x3 + 33)
Use the rule a3 + b3 = (a + b)
(a2 – ab + b2). 4x(x + 3)(x2 – x 3 + 32)
4x(x + 3)(x2 – 3x + 9)
Holt Algebra 2
6-4 Factoring Polynomials
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 2 + 22] Use the rule a3 – b3 =
(a – b) (a2 + ab + b2).
(5d – 2)(25d2 + 10d + 4)
Holt Algebra 2
6-4 Factoring Polynomials
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)
Holt Algebra 2
6-4 Factoring Polynomials
Check It Out! Example 3b
Factor the expression.
2x5 – 16x2
Factor out the GCF, 2x2. 2x2(x3 – 8)
Rewrite as the difference of
cubes. 2x2(x3 – 23)
Use the rule a3 – b3 = (a – b)
(a2 + ab + b2). 2x2(x – 2)(x2 + x 2 + 22)
2x2(x – 2)(x2 + 2x + 4)
Holt Algebra 2
6-4 Factoring Polynomials
Example 4: Geometry Application
The volume of a plastic storage box is modeled by the function V(x) = x3 + 6x2 + 3x – 10. Identify the values of x for which V(x) = 0, then use the graph to factor V(x).
V(x) has three real zeros at x = –5, x = –2, and x = 1. If the model is accurate, the box will have no volume if x = –5, x = –2, or x = 1.
Holt Algebra 2
6-4 Factoring Polynomials
Use synthetic division to
factor the polynomial.
1 6 3 –10 1
1
1 0
V(x)= (x – 1)(x2 + 7x + 10)
10 7
10 7
Write V(x) as a product.
V(x)= (x – 1)(x + 2)(x + 5) Factor the quadratic.
Example 4 Continued
One corresponding factor is (x – 1).
Holt Algebra 2
6-4 Factoring Polynomials
Check It Out! Example 4
The volume of a rectangular prism is modeled by the function V(x) = x3 – 8x2 + 19x – 12, which is graphed below. Identify the values of x for which V(x) = 0, then use the graph to factor V(x).
V(x) has three real zeros at x = 1, x = 3, and x = 4. If the model is accurate, the box will have no volume if x = 1, x = 3, or x = 4.
Holt Algebra 2
6-4 Factoring Polynomials
Use synthetic division to
factor the polynomial.
1 –8 19 –12 1
1
1 0
V(x)= (x – 1)(x2 – 7x + 12)
12 –7
12 –7
Write V(x) as a product.
V(x)= (x – 1)(x – 3)(x – 4) Factor the quadratic.
Check It Out! Example 4 Continued
One corresponding factor is (x – 1).
Holt Algebra 2
6-4 Factoring Polynomials
4. x3 + 3x2 – 28x – 60
Lesson Quiz
2. x + 2; P(x) = x3 + 2x2 – x – 2
1. x – 1; P(x) = 3x2 – 2x + 5
8(2p – q)(4p2 + 2pq + q2)
(x + 3)(x + 3)(x – 3) 3. x3 + 3x2 – 9x – 27
P(1) ≠ 0, so x – 1 is not a factor of P(x).
P(2) = 0, so x + 2 is a factor of P(x).
4. 64p3 – 8q3
(x + 6)(x – 5)(x + 2)