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Holt McDougal Algebra 2 Factoring Polynomials How do we use the Factor Theorem to determine factors of a polynomial? How do we factor the sum and difference.

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Presentation on theme: "Holt McDougal Algebra 2 Factoring Polynomials How do we use the Factor Theorem to determine factors of a polynomial? How do we factor the sum and difference."— Presentation transcript:

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2 Holt McDougal Algebra 2 Factoring Polynomials How do we use the Factor Theorem to determine factors of a polynomial? How do we factor the sum and difference of two cubes

3 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.

4 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 (x + 1); (x 2 – 3x + 1) Find P(–1) by synthetic substitution. P(–1) = 5 P(–1) ≠ 0, so (x + 1) is not a factor of P(x) = x 2 – 3x + 1.

5 Holt McDougal Algebra 2 Factoring Polynomials Determine whether the given binomial is a factor of the polynomial P(x). Example 2: Determining Whether a Linear Binomial is a Factor (x + 2); (3x 4 + 6x 3 – 5x – 10) Find P(–2) by synthetic substitution. P(–2) = 0, so (x + 2) is a factor of P(x) = 3x 4 + 6x 3 – 5x – 10.

6 Holt McDougal Algebra 2 Factoring Polynomials Determine whether the given binomial is a factor of the polynomial P(x). Example 3: Determining Whether a Linear Binomial is a Factor (x + 2); (4x 2 – 2x + 5) Find P(–2) by synthetic substitution. P(–2) = 25 P(–2) ≠ 0, so (x + 2) is not a factor of P(x) = 4x 2 – 2x + 5.

7 Holt McDougal Algebra 2 Factoring Polynomials Determine whether the given binomial is a factor of the polynomial P(x). Example 4: Determining Whether a Linear Binomial is a Factor (3x – 6); (3x 4 – 6x 3 + 6x 2 + 3x – 30) Find P(2) by synthetic substitution. P(–2) = 0, so (x + 2) is a factor of P(x) = 3x 4 + 6x 3 – 5x – 10. Divide everything by 3 (x – 2); (x 4 – 2x 3 + 2x 2 + x – 10)

8 Holt McDougal Algebra 2 Factoring Polynomials BinomialTrinomial4 or more terms GCF Difference of Two Squares Perfect Square Trinomial Two Binomials a = 1 (Shortcut) Two Binomials a ≠ 1 (Cross) Grouping Methods of Factoring Sum of Two Cubes Difference of Two Cubes

9 Holt McDougal Algebra 2 Factoring Polynomials Lesson 3.5 Practice A


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