Essential Questions

1. Essential Questions • How do we use the Factor Theorem to determine factors of a polynomial? • How do we factor the sum and difference of two cubes

2. 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.

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

4. Example 2: Determining Whether a Linear Binomial is a Factor Determine whether the given binomial is a factor of the polynomial P(x). (x + 2); (3x4 + 6x3 – 5x – 10) Find P(–2) by synthetic substitution. P(–2) = 0, so (x + 2) is a factor of P(x) = 3x4 + 6x3 – 5x – 10.

5. Example 3: Determining Whether a Linear Binomial is a Factor Determine whether the given binomial is a factor of the polynomial P(x). (x + 2); (4x2 – 2x + 5) Find P(–2) by synthetic substitution. P(–2) = 25 P(–2) ≠ 0, so (x + 2) is not a factor of P(x) = 4x2 – 2x + 5.

6. Example 4: Determining Whether a Linear Binomial is a Factor Determine whether the given binomial is a factor of the polynomial P(x). Divide everything by 3 (3x – 6); (3x4 – 6x3 + 6x2 + 3x – 30) (x – 2); (x4 – 2x3 + 2x2 + x – 10) Find P(2) by synthetic substitution. P(–2) = 0, so (x + 2) is a factor of P(x) = 3x4 + 6x3 – 5x – 10.

7. Methods of Factoring GCF GCF GCF Difference of Two Squares Perfect Square Trinomial Grouping Two Binomials a = 1 (Shortcut) Sum of Two Cubes Two Binomials a ≠ 1 (Cross) Difference of Two Cubes

8. Lesson 3.5 Practice A