Objectives Factor the sum and difference of two cubes.

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Presentation transcript:

Objectives Factor the sum and difference of two cubes. Factor by grouping.

Example 1: 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 2 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 3 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 4: 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)

Example 5: 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)

Check It Out! Example 6 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 7 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)

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.

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 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.

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