 # Warm Up Factor each expression. 1. 3x – 6y 3(x – 2y) 2. a2 – b2

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Warm Up Factor each expression. 1. 3x – 6y 3(x – 2y) 2. a2 – b2
(a + b)(a – b) Find each product. 3. (x – 1)(x + 3) x2 + 2x – 3 4. (a + 1)(a2 + 1) a3 + a2 + a + 1

Objectives Use the Factor Theorem to determine factors of a polynomial. Factor the sum and difference of two cubes.

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.

Example 1: Determining Whether a Linear Binomial is a Factor
Determine whether the given binomial is a factor of the polynomial P(x). A. (x + 1); (x2 – 3x + 1) B. (x + 2); (3x4 + 6x3 – 5x – 10) Find P(–1) by synthetic substitution. Find P(–2) by synthetic substitution. –1 1 –3 1 –1 4 –2 –5 –10 1 –4 5 –6 10 3 –5 P(–1) = 5 P(–1) ≠ 0, so (x + 1) is not a factor of P(x) = x2 – 3x + 1. P(–2) = 0, so (x + 2) is a factor of P(x) = 3x4 + 6x3 – 5x – 10.

Find P(–2) by synthetic substitution.
Check It Out! Example 1 Determine whether the given binomial is a factor of the polynomial P(x). a. (x + 2); (4x2 – 2x + 5) b. (3x – 6); (3x4 – 6x3 + 6x2 + 3x – 30) Find P(–2) by synthetic substitution. Divide the polynomial by 3, then find P(2) by synthetic substitution. –2 4 –2 5 –8 20 2 1 – –10 4 –10 25 2 4 10 P(–2) = 25 1 2 5 P(–2) ≠ 0, so (x + 2) is not a factor of P(x) = 4x2 – 2x + 5. P(2) = 0, so (3x – 6) is a factor of P(x) = 3x4 – 6x3 + 6x2 + 3x – 30.

Example 2: Factoring by Grouping
Factor: x3 – x2 – 25x + 25. (x3 – x2) + (–25x + 25) Group terms. 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)

Check It Out! Example 2a Factor: x3 – 2x2 – 9x + 18. (x3 – 2x2) + (–9x + 18) Group terms. 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)

Check It Out! Example 2b Factor: 2x3 + x2 + 8x + 4. (2x3 + x2) + (8x + 4) Group terms. 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)

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.

Example 3A: Factoring the Sum or Difference of Two Cubes
Factor the expression. 4x x 4x(x3 + 27) Factor out the GCF, 4x. 4x(x3 + 33) Rewrite as the sum of cubes. Use the rule a3 + b3 = (a + b)  (a2 – ab + b2). 4x(x + 3)(x2 – x  ) 4x(x + 3)(x2 – 3x + 9)

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  ] Use the rule a3 – b3 = (a – b)  (a2 + ab + b2). (5d – 2)(25d2 + 10d + 4)

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)

Check It Out! Example 3b Factor the expression. 2x5 – 16x2 2x2(x3 – 8) Factor out the GCF, 2x2. Rewrite as the difference of cubes. 2x2(x3 – 23) Use the rule a3 – b3 = (a – b)  (a2 + ab + b2). 2x2(x – 2)(x2 + x  ) 2x2(x – 2)(x2 + 2x + 4)

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