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Chapter 6 Factoring Polynomials

Chapter 6 Factoring Polynomials. Section 5 Summary of Factoring Techniques. Section 6.5 Objectives. 1 Factor Polynomials Completely. Steps for Factoring. Steps for Factoring Step 1: Is there a greatest common factor? If so, factor out the GCF. Step 2: Count the number of terms.

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Chapter 6 Factoring Polynomials

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  1. Chapter 6 Factoring Polynomials Section 5 Summary ofFactoring Techniques

  2. Section 6.5 Objectives 1 Factor Polynomials Completely

  3. Steps for Factoring Steps for Factoring Step 1: Is there a greatest common factor? If so, factor out the GCF. Step 2: Count the number of terms. Step 3: (a) 2 terms (Binomials) • Is it the difference of two squares? If so, • a2 – b2 = (a – b)(a + b) • Is it the sum of two squares? If so, stop! • Is it the difference of two cubes? If so, • a3 – b3 = (a – b)(a2 + ab + b2) • Is it the sum of two cubes? If so, • a3 + b3 = (a + b)(a2 – ab + b2) Continued.

  4. Steps for Factoring Steps for Factoring (Continued) (b) 3 terms (Trinomials) • Is it a perfect square trinomial? If so, • a2 + 2ab + b2 = (a + b)2 or a2 – 2ab + b2 = (a – b)2 • Is the coefficient of the square term 1? If so, • x2 + bx + c = (x + m)(x + n) where mn = c and m + n = b • Is the coefficient of the square term  1? If so, • a. Use factoring by grouping • b. Use trial and error (c) 4 terms • Use factoring by grouping Step 4: Check your work by multiplying out the factors.

  5. Factoring Completely Example: Factor completely: 6x2 6x  36 6x2 6x  36 = 6(x2  x  6) Factor out the GCF, 6. 3 terms, the coefficient of x2 is 1. = 6(x  3)( x + 2) Factor. – 3(2) = – 6 and – 3 + 2 = – 1 Check: 6(x  3)( x + 2) = 6(x2 – x – 6) = 6x2 – 6x – 36 

  6. Factoring Completely Example: Factor completely: 1  16x4 1 and 16x4 are both perfect squares so we have the difference of two squares. 1  16x4 = [12  (4x2)2] = (1  4x2)(1 + 4x2) Factor. a = 1, b = 4x2 This is also a difference of two squares. = (1  2x)(1 + 2x)(1 + 4x2) Factor. Check: (1  2x)(1 + 2x)(1 + 4x2) = (1  4x2)(1 + 4x2) = 1  16x4 

  7. Factoring Completely Example: Factor completely: 80w3 10 80w3 10 = 10(8w3  1) Factor out the GCF, 10. 8w3 and 1 are both perfect cubes, so we have the difference of two cubes. = 10[(2w)3  13] a = 2w, b = 1. = 10(2w  1)(4w2 + 2w + 1) Factor. Check: 10(2w  1)(4w2 + 2w + 1) = 10(8w3  1) = 80w3 10 

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