holt mcdougal algebra 2 6-4 factoring polynomials 6-4 factoring polynomials holt algebra 2 warm up...
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Holt McDougal Algebra 2
6-4 Factoring Polynomials6-4 Factoring Polynomials
Holt Algebra 2
Warm UpWarm Up
Lesson PresentationLesson Presentation
Lesson QuizLesson Quiz
Holt McDougal Algebra 2
Holt McDougal Algebra 2
6-4 Factoring Polynomials
Warm UpFactor 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 McDougal 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 McDougal 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 McDougal 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
1000
–500
P(–2) = 0, so (x + 2) is a factor of P(x) = 3x4 + 6x3 – 5x – 10.
Holt McDougal 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
1040
520
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 McDougal 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 McDougal 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 McDougal 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 McDougal 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 McDougal 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 McDougal 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 McDougal 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 McDougal 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 McDougal 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 McDougal 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 McDougal 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 McDougal Algebra 2
6-4 Factoring Polynomials
Example 4: Geometry ApplicationThe 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 McDougal 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)
107
107
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 McDougal 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 McDougal 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 McDougal 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)