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

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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.

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 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)

Lesson Quiz 1. x – 1; P(x) = 3x2 – 2x + 5 P(1) ≠ 0, so x – 1 is not a factor of P(x). 2. x + 2; P(x) = x3 + 2x2 – x – 2 P(2) = 0, so x + 2 is a factor of P(x). 3. x3 + 3x2 – 9x – 27 (x + 3)(x + 3)(x – 3) 4. x3 + 3x2 – 28x – 60 (x + 6)(x – 5)(x + 2) 4. 64p3 – 8q3 8(2p – q)(4p2 + 2pq + q2)

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