Copyright © 2015, 2011, 2007 Pearson Education, Inc. 1 1 Chapter 6 Factoring

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 2 Factoring 6.1Greatest Common Factor and Factoring by Grouping 6.2Factoring Trinomials 6.3Factoring Special Products and Factoring Strategies 6.4Solving Equations by Factoring CHAPTER 6

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 3 Factoring Trinomials 1.Factor trinomials of the form x 2 + bx + c. 2.Factor trinomials of the form ax 2 + bx + c, where a  1, by trial. 3.Factor trinomials of the form ax 2 + bx + c, where a  1, by grouping. 4.Factor trinomials of the form ax 2 + bx + c using substitution. 6.2

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 4 First we consider trinomials in the form x 2 + bx + c in which the coefficient of the squared term is 1. If the trinomial is factorable, we have two binomial factors. (x + 2)(x + 3) = x 2 + 3x + 2x + 6 = x 2 + 5x + 6 Sum of 2 and 3 Product of 2 and 3 F O I L

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 5 Factoring x 2 + bx + c To factor a trinomial of the form x 2 + bx + c, 1. Find two numbers with a product equal to c and a sum equal to b. 2. The factored trinomial will have the form: (x + ) (x + ), where the second terms are the numbers found in Step 1. Note: The signs of b and c may cause one or both signs in the binomial factors to be minus signs.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 6 Example Factor. x 2 – 6x + 8 Solution We must find a pair of numbers whose product is 8 and whose sum is –6. Note that if two numbers have a positive product and negative sum, they must both be negative. List the possible products and sums. ProductSum (–1)(–8) = 8–1 + (–8) = –9 (–2)(–4) = 8–2 + (–4) = –6 This is the correct combination.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 7 continued Answer Check Multiply the factors to verify that their product is the original polynomial. x 2 – 6x + 8 = (x +(– 2))(x+( – 4)) (x – 2)(x – 4) = x 2 – 4x – 2x + 8= x 2 – 6x + 8 It checks. = (x – 2)(x – 4)

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 8 Example Factor. x 2 + 2x – 24 Solution: We must find a pair of numbers whose product is –24 and whose sum is 2. Because the product is negative, the two numbers have different signs. ProductSum (–2)(12) =  24 – = 10 (–4)(6) =  24 –4 + 6 = 2 This is the correct combination.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 9 continued Answer Check We can check by multiplying the binomial factors to see if their product is the original polynomial. x 2 + 2x – 24 = (x – 4)(x + 6) (x – 4)(x + 6) = x 2 + 6x – 4x – 24 = x 2 + 2x – 24 Multiply the factors using FOIL. The product is the original polynomial.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 10 Example Factor. 3x 4 – 18x x 2 Solution First, factor out the GCF, 3x 2. 3x 4 – 18x x 2 = 3x 2 (x 2 – 6x + 8) = 3x 2 (x – 4)(x – 2)

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 11 Example Factor. x 4 + 2x 3 + 5x 2 Solution First, factor out the GCF, x 2. x 2 ( x 2 + 2x + 5) Now try to factor the trinomial to two binomials. We must find a pair of numbers whose product is 5 and whose sum is 2. ProductSum (1)(5) = = 6

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 12 continued Factor. x 4 + 2x 3 + 5x 2 The trinomial x 2 + 2x + 5 has no binomial factors with integer terms, so x 2 (x 2 + 2x + 5) is the final factored form. There is no combination of factors whose product is 5 and sum is 2.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 13 Example Factor. Solution First, factor out the GCF, 2xy.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 14 Example Factor. a 2 – ab – 20b 2 Solution We must find a pair of terms whose product is  20b 2 and whose sum is –1b. Answer a 2 – ab – 20b 2 = (a – 5b)(a + 4b) ProductSum (–1b)(20b) = –20b 2 –1b+ 20b= 19b (1b)(–20b) = – 20b 2 1b+ (–20b) = –19b (2b)(–10b) = – 20b 2 2 b+ (–10b) = –8b (–2b)(10b) = – 20b 2 –2b + 10b = 8b (4b)(–5b) = – 20b 2 4b + (–5b) = –1b (–4b)(5b) = – 20b 2 –4b + 5b = 1b This is the correct combination.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 15 Example Factor. 4xy xy 2 – 72xy Solution This trinomial has a monomial GCF, 4xy. 4xy xy 2 – 72xy = 4xy(y 2 + 3y – 18) We must find a pair of numbers whose product is –18 and whose sum is 3. = 4xy(y – 3)(y + 6)

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 16 Example Factor. Solution The first terms must multiply to equal 6x 2. These could be x and 6x, or 2x and 3x. The last terms must multiply to equal –5. Because –5 is negative, the last terms in the binomials must have different signs. This factor pair must be 1 and 5.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 17 continued Now we multiply binomials with various combinations of these first and last terms until we find a combination whose inner and outer products combine to equal 13x. Correct combination. Incorrect combinations. Answer

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 18 Factoring by Trial and Error To factor a trinomial of the form ax 2 + bx + c, where a ≠ 1, by trial and error: 1. Look for a monomial GCF in all of the terms. If there is one, factor it out. 2. Write a pair of first terms whose product is ax 2. ax 2

Copyright © 2015, 2011, 2007 Pearson Education, Inc Write a pair of last terms whose product is c. 4. Verify that the sum of the inner and outer products is bx (the middle term of the trinomial). c + Outer bx Inner

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 20 If the sum of the inner and outer products is not bx, then try the following: a. Exchange the last terms of the binomials from step 3; then repeat step 4. b. For each additional pair of last terms, repeat steps 3 and 4. c. For each additional pair of first terms, repeat steps 2 – 4.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 21 Example Factor. Solution The last terms must multiply to equal 2. Because 2 is a prime number, its factors are 1 and 2. There is no common monomial factor. The first terms must multiply to equal 6x 2. These could be x and 6x, or 3x and 2x.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 22 continued Now find a combination whose inner and outer products combine to equal –x. Correct combination. Answer

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 23 Example Factor. Solution The last terms must multiply to equal 3. Because 3 is a prime number, its factors are 1 and 3. Now we factor the trinomial within the parentheses. The first terms must multiply to equal 7x 2. These could be x and 7x. Factor out the monomial GCF, 3x.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 24 continued Now find a combination whose inner and outer products combine to equal –20x. Correct combination. Answer

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 25 Factoring ax 2 + bx + c, Where a ≠ 1, by Grouping To factor a trinomial of the form ax 2 + bx + c, where a ≠ 1, by grouping, 1. Look for a monomial GCF in all of the terms. If there is one, factor it out. 2. Find two factors of the product ac whose sum is b. 3. Write a four-term polynomial in which bx is written as the sum of two like terms whose coefficients are the two factors you found in step Factor by grouping.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 26 Example Factor. Solution For this trinomial, a = 2, b = –15, and c = 7, so ac = (2)(7) = 14. We must find two factors of 14 whose sum is –15. Since 14 is positive and –15 is negative, the two factors must be negative. Factors of acSum of Factors of ac (–2)(–7) = 14–2 + (–7) = –9 (–1)(–14) = 14–1 + (– 14) = –15 Correct

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 27 continued –15x–x – 14x Write -15x as –x – 14. Factor x out 2x 2 – x, factor -7 out of -14x+7. Factor out of 2x – 1.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 28 Example Factor. Solution Notice that there is a GCF, 2x 2, that we can factor out. 2x 2 (6x 2 + 7x – 3) Now we factor the trinomial within the parentheses. a = 6, c =  3; ac = (6)(  3) =  18 Find two factors of  18 whose sum is 7. Because the product is negative, the two factors will have different signs.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 29 continued Factor. Solution Factors of acSum of Factors of ac (  1)(18) =  18  = 17 (  2)(9) =  18  = 7 (  3)(6) =  18  = 3 Correct 2x 2 (6x 2 + 7x – 3) Now write 7x as  2x + 9x and then factor by grouping.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 30 continued +7x  2x + 9x

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 31 Example Factor using substitution. Solution If we substitute another variable, like u, for y – 5, we can see that the polynomial is actually in the form ax 2 + bx + c. Substitute u for y – 5. We can factor by grouping.

Copyright © 2015, 2011, 2007 Pearson Education, Inc. 32 continued Substitute y – 5 for u. Remember that u was substituted for y – 5, so we must substitute y – 5 back for u to have the factored form of our original polynomial. This resulting factored form can be simplified. Combine like terms. Distribute.