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6-4. Factoring Polynomials. Warm Up. Lesson Presentation. Lesson Quiz. Holt Algebra 2. 8 x 2 – 10 x + 20 –. x 2 – 2 x + 1 +. 33. 3. x + 2. x + 2. Warm Up. 1. Divide by using long division. (8 x 3 + 6 x 2 + 7) ÷ ( x + 2).

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  1. 6-4 Factoring Polynomials Warm Up Lesson Presentation Lesson Quiz Holt Algebra 2

  2. 8x2– 10x + 20 – x2– 2x + 1 + 33 3 x + 2 x + 2 Warm Up 1. Divide by using long division.(8x3 + 6x2 + 7) ÷ (x + 2) 2. Divide by using synthetic division. (x3– 3x + 5) ÷ (x+ 2) 3. Use synthetic substitution to evaluate P(x) = x3 + 3x2 – 6 for x = 5 and x = –1. 194; –4

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

  4. 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.

  5. 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 3 6 0 –5 –10 1 –4 5 –6 0 0 10 3 0 0 –5 0 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.

  6. Check It Out! Example 1 Determine whether the given binomial is a factor of the polynomial P(x). b. (3x – 6); (3x4 – 6x3 + 6x2 + 3x – 30) a. (x + 2); (4x2 – 2x + 5) 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 –2 2 1 –10 4 –10 25 2 0 4 10 1 0 2 5 0 P(–2) = 25 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.

  7. 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)

  8. 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)

  9. 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.

  10. 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. Use the rule a3 + b3 = (a + b)  (a2– ab + b2). 4x(x + 3)(x2–x3+32) 4x(x + 3)(x2 – 3x + 9)

  11. 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 2 + 22] Use the rule a3– b3 = (a – b)  (a2+ ab + b2). (5d – 2)(25d2 + 10d + 4)

  12. 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2+22) 2x2(x – 2)(x2 + 2x + 4)

  13. 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.

  14. Example 4 Continued One corresponding factor is (x – 1). 1 1 6 3 –10 Use synthetic division to factor the polynomial. 1 7 10 1 0 7 10 V(x)= (x – 1)(x2 + 7x + 10) Write V(x) as a product. V(x)= (x – 1)(x+ 2)(x + 5) Factor the quadratic.

  15. 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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