Holt McDougal Algebra 2 Factoring Polynomials Warm Up Factor each expression. 1. 3x – 6y 2. a 2...
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Transcript of Holt McDougal Algebra 2 Factoring Polynomials Warm Up Factor each expression. 1. 3x – 6y 2. a 2...
Holt McDougal Algebra 2
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
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
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
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
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
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
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
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
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
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
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
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
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)