1. 5.7-5.9 More about Factoring Trinomials

2. Factoring a trinomial of the form ax 2 +bx+c To factor ax 2 +bx+c when a≠1 find the integers k,l,m,n such that ax 2 +bx+c=(kx+m)(lx+n) = klx 2 +(kn+lm)x+mn Therefore k and l must be factors of a m and n must be factors of c The goal is to find a combination of factors of a and c such that the outer and inner products add up to the middle term bx.

3. Example 3x 2 + x -4 The possible factors for the first term are 3,1 or 1,3. The possible factors of the last term are 2, -2; -2, 2; 4, -1; and -4, 1. Here are all the possible factors and what the middle term equals (3x -2) (x + 2) = 4x for the middle term (3x + 2) (x – 2) = -4x for the middle term (3x -1) (x + 4) = 11x for the middle term (3x-4) (x + 1) = -x for the middle term (3x + 1) (x-4) = -11 for the middle term (3x +4) (x – 1) = x for the middle term ANSWER

4. Examples: 1) 2) 3)

5. Examples: Factoring Completely 4) 5)

6. Factoring by grouping Factor: 2x 2 + x -15 In order to ‘group’ this polynomial, we actually have to ‘break out’ the middle term. In order to figure out what coefficients to use when we break out the middle term, we need to choose factors of the product ac that add up to b. In the example above the product of ac is -30 (-15*2). The factors of -30 are +/- 1, 30, 5, 6, 3 10, 2 and 15. Looking at this factors, we see that -5 + 6 will equal one. So now we can write out this polynomial like this: 2x 2 + 6x-5x -15. 2x(x + 3) -5(x+3) (2x-5)(x+3)

7. Factoring by grouping If necessary, write the trinomial in standard form Choose factors of the product ac that add up to b Use these factors to rewrite the middle term as a sum. Group factor and solve.

8. Examples by grouping: 1) 2)