5 4 Factoring Trinomials n Factoring Trinomials of

5. 4 Factoring Trinomials n Factoring Trinomials of the Type x 2 + bx + c n Factoring Trinomials of the Type ax 2 + bx + c, Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Factoring Trinomials of the Type x 2 + bx + c When trying to factor trinomials of the type x 2 + bx + c, we can use a trial-anderror procedure. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 2

Constant Term Positive Recall the FOIL method of multiplying two binomials: F O I L (x + 2)(x + 5) = x 2 + 5 x + 2 x + 10 = x 2 + 7 x + 10 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 3

To factor x 2 + 7 x + 10, we think of FOIL: The first term, x 2, is the product of the First terms of two binomial factors, so the first term in each binomial must be x. The challenge is to find two numbers p and q such that x 2 + 7 x + 10 = (x + p)(x + q) = x 2 + qx + pq Thus the numbers p and q must be selected so that their product is 10 and their sum is 7. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 4

In this case, we know from a previous slide that these numbers are 2 and 5. The factorization is (x + 2)(x + 5), or (x + 5)(x + 2). Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 5

Example Write an equivalent expression by factoring: x 2 + 10 x + 21. Solution We search for factors of 21 whose sum is 10. Pair of Factors 3, 7 1, 21 Sum of Factors 10 23 The numbers we need are 3 and 7. The factorization is thus (x + 7)(x + 3). Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 6

Example Write an equivalent expression by factoring: x 2 – 12 x + 20. Solution We search for factors of 20 whose sum is – 12. Pair of Factors Sum of Factors -4, -5 -2, -10 -1, -20 -9 -12 -21 The numbers we need are -2 and -10. The factorization is thus (x – 2)(x – 10). Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 7

Constant Term Negative When the constant term of a trinomial is negative, we look for one negative factor and one positive factor. The sum of the factors must still be the coefficient of the middle term. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 8

Example Write an equivalent expression by factoring: x 2 – x – 20. Solution We search for factors of – 20 whose sum is – 1. Pair of Factors 4, -5 2, -10 1, -20 Sum of Factors -1 -8 -19 The numbers we need are 4 and -5. The factorization is thus (x – 5)(x + 4). Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 9

Some polynomials are not factorable using integers. 2 + 2 x + 4 Factor: x Example Solution There are no factors of 4 whose sum is 2. This trinomial is not factorable into binomials with integer coefficients. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 10

Tips for Factoring x 2 + bx + c 1. If necessary, rewrite the trinomial in descending order. 2. Find a pair of factors that have c as their product and b as their sum. Remember the following: n If c is positive, its factors will have the same sign as b. n If c is negative, one factor will be positive and the other will be negative. Select the factors such that the factor with the larger absolute value is the factor with the same sign as b. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 11

Tips for Factoring x 2 + bx + c n If the sum of the two factors is the opposite of b, changing the signs of both factors will give the desired factors whose sum is b. 3. Check the result by multiplying the binomials. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 12

Example Factor: x 2 – 3 xy – 28 y 2. Solution We look for numbers p and q such that x 2 – 3 xy – 28 y 2 = (x + py)(x + qy) Approach it as though it is x 2 – 3 x – 28. We look for factors that multiply to make – 28 and whose sum is – 3. Those factors are – 7 and 4. Thus, x 2 – 3 xy – 28 y 2 = (x – 7 y)(x + 4 y) The check is left to the student. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 13

Factoring Trinomials of the 2 Type ax + bx + c (a not 1) Now we look at trinomials in which the leading coefficient is not 1. We consider two methods. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 14

Method 1: Reversing FOIL We first consider the FOIL method for factoring trinomials of the type ax 2 + bx +c, where Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 15

Consider the following multiplication. F O I L 23 x + 12 (2 x + 3)(5 x + 4) = 10 x 2 + 8 x + 15 x + 12 = 10 x 2 + To reverse what we did we look for two binomials whose product is this trinomial. The product of the First terms must be 10 x 2. The product of the Outer terms plus the product of the Inner terms must be 23 x. The product of the Last terms must be 12. In general, finding such an answer involves trial and error. We use the following method. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 16

To Factor ax 2 + bx = c by Reversing FOIL 1. Factor out the largest common factor, if one exists. Here we assume that none does. 2. Find two First terms whose product is ax 2: ( x + ) = ax 2 + bx + c 3. Find two Last terms whose product is c: ( x + ) = ax 2 + bx + c 4. Repeat steps (2) and (3) until a combination is found for which the sum of the Outer and Inner products is bx: ( x + ) = ax 2 + bx + c Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 17

Example Factor: 2 x 2 + 7 x + 6 Solution 1. Observe that there are no common factors (other than 1 or -1). 2. To factor the first term it must be 2 x times x. (2 x + )(x + ) 3. The constant term 6, can be factored as (6)(1), (3)(2), (-6)(-1), and (-3)(-2). 4. Find a pair for which the sum of the Outer and Inner products is 7 x. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 18

Each possibility should be checked by multiplying. (2 x + 1)(x + 6) O + I = 12 x + x = 13 x This is not correct. Try again. (2 x + 3)(x + 2) O + I = 4 x + 3 x = 7 x This is the desired factorization. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 19

Tips for Factoring with FOIL 1. If the largest common factor has been factored out of the original trinomial, then no binomial factor can have a common factor (except 1, -1). 2. If a and c are both positive, then the signs of the factors will be the same as the sign of b. 3. When a possible factoring produces the opposite of the desired middle term, reverse the signs of the constants in the factors. 4. Keep track of those possibilities that you have tried and those you that you have not. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 20

Method 2: The Grouping Method The second method for factoring trinomials of the type ax 2 + bx + c, is known as the grouping method. It involves not only trial and error and FOIL but also factoring by grouping. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 21

Example Factor: 4 x 2 + 16 x + 15. First, multiply the leading coefficient, 4, and the constant, 15, to get 60. Then find a factorization of 60 in which the sum of the factors is the coefficient of the middle term: 16 (in our case 6 and 10). The middle term is then split into the sum or difference using these factors. Then factor by grouping. 4 x 2 + 16 x + 15 = 4 x 2 + 6 x + 10 x + 15 = 2 x(2 x + 3) + 5(2 x + 3) = (2 x + 3)(2 x + 5) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 22

To Factor ax 2 +bx + c Using Grouping 1. Make sure that any common factors have been factored out. 2. Multiply the leading coefficient a and the constant c. 3. Find a pair of factors, p and q, so that pq = ac and p + q = b. 4. Rewrite the trinomial’s middle term, bx, as px + qx. 5. Factor by grouping. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 23

Example Factor: 6 x 2 + 5 x – 6. Solution We look for factors of – 36 that add to 5. The factors 9 and – 4 are the factors we seek. 6 x 2 + 5 x – 6 = 6 x 2 + 9 x – 4 x – 6 = 3 x(2 x + 3) – 2(2 x + 3) = (2 x + 3)(3 x – 2) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 5 - 24