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6-4 6-4 Factoring Polynomials Warm Up Lesson Presentation Lesson Quiz Holt Algebra Holt Algebra 22 6-4 Factoring Polynomials Warm Up 1. Divide by using long division. (8x3 + 6x2 + 7) ÷ (x + 2) 8x2 – 10x + 20 – 33 x+2 2. Divide by using synthetic division. 3 2 (x3 – 3x + 5) ÷ (x + 2) x – 2x + 1 + x + 2 3. Use synthetic substitution to evaluate 194; –4 3 2 P(x) = x + 3x – 6 for x = 5 and x = –1. Holt Algebra 2 6-4 Factoring Polynomials Objectives Use the Factor Theorem to determine factors of a polynomial. Factor the sum and difference of two cubes. Holt Algebra 2 6-4 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 Algebra 2 6-4 Factoring Polynomials 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) Find P(–1) by synthetic substitution. –1 1 –3 1 1 –1 –4 4 5 P(–1) = 5 P(–1) ≠ 0, so (x + 1) is not a factor of P(x) = x2 – 3x + 1. Holt Algebra 2 B. (x + 2); (3x4 + 6x3 – 5x – 10) Find P(–2) by synthetic substitution. –2 3 6 0 –5 –10 –6 0 0 3 0 0 –5 10 0 P(–2) = 0, so (x + 2) is a factor of P(x) = 3x4 + 6x3 – 5x – 10. 6-4 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. –2 4 –2 5 4 –8 20 –10 25 P(–2) = 25 P(–2) ≠ 0, so (x + 2) is not a factor of P(x) = 4x2 – 2x + 5. Holt Algebra 2 b. (3x – 6); (3x4 – 6x3 + 6x2 + 3x – 30) Divide the polynomial by 3, then find P(2) by synthetic substitution. 2 1 –2 2 1 –10 1 2 0 4 10 0 2 5 0 P(2) = 0, so (3x – 6) is a factor of P(x) = 3x4 – 6x3 + 6x2 + 3x – 30. 6-4 Factoring Polynomials Example 2: Factoring by Grouping Factor: x3 – x2 – 25x + 25. (x3 – x2) + (–25x + 25) x2(x – 1) – 25(x – 1) Group terms. Factor common monomials from each group. (x – 1)(x2 – 25) Factor out the common binomial (x – 1). (x – 1)(x – 5)(x + 5) Factor the difference of squares. Holt Algebra 2 6-4 Factoring Polynomials Check It Out! Example 2a Factor: x3 – 2x2 – 9x + 18. (x3 – 2x2) + (–9x + 18) x2(x – 2) – 9(x – 2) Group terms. Factor common monomials from each group. (x – 2)(x2 – 9) Factor out the common binomial (x – 2). (x – 2)(x – 3)(x + 3) Factor the difference of squares. Holt Algebra 2 6-4 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 Algebra 2 6-4 Factoring Polynomials Example 3A: Factoring the Sum or Difference of Two Cubes Factor the expression. 4x4 + 108x 4x(x3 + 27) Factor out the GCF, 4x. 4x(x3 + 33) Rewrite as the sum of cubes. 4x(x + 3)(x2 –x3+ 32) 4x(x + 3)(x2 – 3x + 9) Holt Algebra 2 Use the rule a3 + b3 = (a + b) (a2 – ab + b2). 6-4 Factoring Polynomials Example 3B: Factoring the Sum or Difference of Two Cubes Factor the expression. 125d3 – 8 Rewrite as the difference of cubes. (5d – 2)[(5d)2 + 5d 2 + 22] Use the rule a3 – b3 = (a – b) (a2 + ab + b2). (5d)3 – 23 (5d – 2)(25d2 + 10d + 4) Holt Algebra 2 6-4 Factoring Polynomials 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) 2x2(x – 2)(x2 +x2+ 22) 2x2(x – 2)(x2 + 2x + 4) Holt Algebra 2 Use the rule a3 – b3 = (a – b) (a2 + ab + b2). 6-4 Factoring Polynomials Example 4: Geometry Application The volume of a plastic storage box is modeled by the function V(x) = x3 + 6x2 + 3x – 10. Identify the values of x for which V(x) = 0, then use the graph to factor V(x). V(x) has three real zeros at x = –5, x = –2, and x = 1. If the model is accurate, the box will have no volume if x = –5, x = –2, or x = 1. Holt Algebra 2 6-4 Factoring Polynomials Example 4 Continued One corresponding factor is (x – 1). 1 1 6 3 –10 1 7 10 1 7 10 0 Use synthetic division to factor the polynomial. V(x)= (x – 1)(x2 + 7x + 10) Write V(x) as a product. V(x)= (x – 1)(x + 2)(x + 5) Factor the quadratic. Holt Algebra 2 6-4 Factoring Polynomials 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 Holt Algebra 2 8(2p – q)(4p2 + 2pq + q2)