Factoring Polynomials MPM 2D Factoring Polynomials

A common binomial factor can be factored out of certain expressions. Example: Factor 4x2 – 12x + 20. GCF = 4. = 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) (x + 1) (5 – y) = 5(x + 1) – y(x + 1) Check. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Factor

D.O.T.S. A difference of two squares can be factored using the formula a2 – b2 = (a + b)(a – b). Example: Factor x2 – 9y2. x2 – 9y2 = (x)2 – (3y)2 Write terms as perfect squares. = (x + 3y)(x – 3y) Use the formula. The same method can be used to factor any expression which can be written as a difference of squares. Example: Factor (x + 1)2 – 25y 4. (x + 1)2 – 25y 4 = (x + 1)2 – (5y2)2 = [(x + 1) + (5y2)][(x + 1) – (5y2)] = (x + 1 + 5y2)(x + 1 – 5y2) Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Difference of Squares

Examples: 1. Factor 2xy + 3y – 4x – 6. Some polynomials can be factored by grouping terms to produce a common binomial factor. Examples: 1. Factor 2xy + 3y – 4x – 6. Notice the sign! 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) b – a = – (a – b). = (a – b) (2a – 3c) Factor. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Examples: Factor

Factoring these trinomials is based on reversing the FOIL process. 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. Express the trinomial as a product of two binomials with leading term x and unknown constant terms a and b. x2 + 3x + 2 = (x + a)(x + b) F O I L = x2 + bx + ax + ba Apply FOIL to multiply the binomials. = x2 + (b + a) x + ba Since ab = 2 and a + b = 3, it follows that a = 1 and b = 2. = x2 + (1 + 2) x + 1 · 2 (Product-sum method) Therefore, x2 + 3x + 2 = (x + 1)(x + 2). Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Factor x2 + bx + c

It follows that both a and b are negative. 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. Sum Negative Factors of 15 - 1, - 15 -15 -3, - 5 - 8 x2 – 8x + 15 = (x – 3)(x – 5). Check: = x2 – 8x + 15. (x – 3)(x – 5) = x2 – 5x – 3x + 15 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Factor

two positive factors of 36 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 Sum Positive Factors of 36 whose sum is 13. 1, 36 37 2, 18 20 15 3, 12 4, 9 13 6, 6 12 = (x + 4)(x + 9) x2 + 13x + 36 Check: (x + 4)(x + 9) = x2 + 9x + 4x + 36 = x2 + 13x + 36. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Factor

Example: Factor Completely 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 The GCF is 4x. = 4x(x2 – 10x + 25) Use distributive property to factor out the GCF. = 4x(x – 5)(x – 5) 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. Example: Factor Completely

Factoring Polynomials of the Form ax2 + bx + c 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 2. We need to find factors of 12 1, 12 2, 6 3, 4 1. Find the product of first and last terms is 8 whose sum 3. Rewrite the middle term decomposed into the two numbers 3x2 + 2x + 6x + 4 = (3x2 + 2x) + (6x + 4) 4. Factor by grouping in pairs = 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. Factoring Polynomials of the Form ax2 + bx + c

Example: Factor 4x2 + 8x – 5. 4  5 = -20 1, 20 -2, 10 4, 5 We need to find factors of 20 1, 20 -2, 10 4, 5 is 8 whose difference Rewrite the middle term decomposed into the two numbers 4x2 – 2x + 10x – 5 = (4x2 – 2x) + (10x – 5) Factor by grouping in pairs = 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. Example: Factor