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MPM 2D Factoring Polynomials Example: GCF = 4. Factor 4x2 – 12x + 20. = 4(x2 – 3x + 5) Check the answer. 4(x2 – 3x + 5) = 4x2 – 12x + 20 A common binomial factor can be factored out of certain expressions. Example: Factor the expression 5(x + 1) – y(x + 1). 5(x + 1) – y(x + 1) = (x + 1) (5 – y) Check. (x + 1) (5 – y) = 5(x + 1) – y(x + 1) Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 2 Some polynomials can be factored by grouping terms to produce a common binomial factor. Notice the sign! Examples: 1. Factor 2xy + 3y – 4x – 6. 2xy + 3y – 4x – 6 = (2xy + 3y) – (4x + 6) Group terms. = y (2x + 3) – 2(2x + 3) Factor each pair of terms. = (2x + 3) ( y – 2) Factor out the common binomial. 2. Factor 2a2 + 3bc – 2ab – 3ac. 2a2 + 3bc – 2ab – 3ac = 2a2 – 2ab + 3bc – 3ac Rearrange terms. = (2a2 – 2ab) + (3bc – 3ac) Group terms. = 2a(a – b) + 3c(b – a) Factor. = 2a(a – b) – 3c(a – b) = (a – b) (2a – 3c) Copyright © by Houghton Mifflin Company, Inc. All rights reserved. b – a = – (a – b). Factor. 3 To factor a simple trinomial of the form x2 + bx + c, express the trinomial as the product of two binomials. For example, x2 + 10x + 24 = (x + 4)(x + 6). Factoring these trinomials is based on reversing the FOIL process. Example: Factor x2 + 3x + 2. x2 + 3x + 2 = (x + a)(x + b) Express the trinomial as a product of two binomials with leading term x and unknown constant terms a and b. F O I L = x2 + bx + ax + ba = x2 + (b + a)x + ba = x2 + (1 + 2)x + 1 · 2 Apply FOIL to multiply the binomials. Since ab = 2 and a + b = 3, it follows that a = 1 and b = 2. (Product-sum method) Therefore, x2 + 3x + 2 = (x + 1)(x + 2). Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 4 Example: Factor x2 – 8x + 15. x2 – 8x + 15 = (x + a)(x + b) = x2 + (a + b)x + ab Therefore a + b = -8 and ab = 15. It follows that both a and b are negative. Negative Factors of 15 Sum - 1, - 15 -15 -3, - 5 -8 x2 – 8x + 15 = (x – 3)(x – 5). Check: (x – 3)(x – 5) = x2 – 5x – 3x + 15 = x2 – 8x + 15. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 5 13 + 36 Example: Factor x2 + 13x 36. x2 + 13x + 36 = (x + a)(x + b) = x2 + (a + b)x + ab Therefore a and b are: two positive factors of 36 whose sum is 13. Positive Factors of 36 Sum 1, 36 37 2, 18 3, 12 20 15 4, 9 6, 6 13 12 x2 + 13x + 36 = (x + 4)(x + 9) Check: (x + 4)(x + 9) = x2 + 9x + 4x + 36 = x2 + 13x + 36. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 6 A polynomial is factored completely when it is written as a product of factors that can not be factored further. Example: Factor 4x3 – 40x2 + 100x. 4x3 – 40x2 + 100x = 4x(x2 – 10x + 25) = 4x(x – 5)(x – 5) The GCF is 4x. Use distributive property to factor out the GCF. Factor the trinomial. Check: 4x(x – 5)(x – 5) = 4x(x2 – 5x – 5x + 25) = 4x(x2 – 10x + 25) = 4x3 – 40x2 + 100x Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 7 Factoring complex trinomials of the form ax2 + bx + c, (a 1) can be done by decomposition or cross-check method. Example: Factor 3x2 + 8x + 4. 3 4 = 12 Decomposition Method 1. Find the product of first and last terms 3. Rewrite the middle term decomposed into the two numbers 4. Factor by grouping in pairs 2. We need to find factors of 12 whose sum is 8 1, 12 2, 6 3, 4 3x2 + 2x + 6x + 4 = (3x2 + 2x) + (6x + 4) = x(3x + 2) + 2(3x + 2) = (3x + 2) (x + 2) 3x2 + 8x + 4 = (3x + 2) (x + 2) Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 8 Example: Factor 4x2 + 8x – 5. 4 5 = -20 We need to find factors of 20 whose difference is 8 Rewrite the middle term decomposed into the two numbers Factor by grouping in pairs 1, 20 -2, 10 4, 5 4x2 – 2x + 10x – 5 = (4x2 – 2x) + (10x – 5) = 2x(2x – 1) + 5(2x – 1) = (2x – 1) (2x + 5) 4x2 + 8x – 5 = (2x –1)(2x – 5) Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 9