AC Method

When dealing with a trinomial, we must first make sure that is in descending order. Recall the general form of a quadratic equation; ax2 + bx + c. We break trinomials into 2 cases, trinomials with leading coefficient 1 ie; a = 1, and trinomials with leading coefficient other than 1, a is not equal to 1 Factoring Trinomials when the leading coefficient is 1; a = 1. Look at the factors of c, and find two that add to be b. the below chart may be helpful b+, c+: then both factors are positive b-, c+: then both factors are negative b+, c-: then the larger factor is positive and the smaller factor is negative b-, c-: then the larger factor is negative and the smaller factor is positive Example x2 + 8x +12 b = 8 and c = 12 since both b and c are positive, both factors will be positive the factors of 12 are 1,12 2, 6 3, 4 2 and 6 add to be 8 so our answer is (x + 2)(x + 6) Example x2 - 7x -18 b = -7 and c = -18 since both b and c are negative, the larger factor is negative and the smaller is positive the factors of -18 are 1,-18 2,-9 3, -6 2 and -9 add to be -7 so our answer is (x + 2)(x - 9) Factoring Trinomials when the leading coefficient is not 1; a is not equal to 1. We first multiply a times c, we then look at the factors of ac, and find two that add to be b. We break up the middle term using the two factors. We now will have a polynomial with 4 terms, which means we will use factoring by grouping. the below chart may be helpful b+, ac+: then both factors are positive b-, ac+: then both factors are negative b+, ac-: then the larger factor is positive and the smaller factor is negative b-, ac-: then the larger factor is negative and the smaller factor is positive Example 3x2 + 14x + 8 ac = 24 and b = 14 since both b and ac are positive, both factors will be positive the factors of 24 are 1,24 2, 12 3, 8 4,6 2 and 12 add to be 24 so we use 2 and 12 to break up the middle term 3x2 + 2x + 12x + 8; note: we have not changed the problem since 2x + 12x = 14x Now we can use factoring by grouping (3x2 + 2x) + (12x + 8) x(3x + 2) + 4(3x + 2) and our answer is (3x + 2)(x + 4) Example 4x2 - 11x -15 ac = -60 and b = -11 since both b and ac are negative, the larger factor is negative and the smaller is positive the factors of -60 are 1,-60 2,-30 3, -20 4,-15 5,-12 6,-10 4 and -15 add to be -11 so we use 4 and -15 to break up the middle term 4x2 + 4x - 15x - 15; note: we have not changed the problem since 4x + -15x = 11x Now we can use factoring by grouping (4x2 + 4x) + (-15x - 15) 4x(x + 1) - 15(x + 1) and our answer is (x + 1)(4x - 15) Factor each of the following. 1. x2 + 2x - 15 2. 3x3 - 21x2 + 36x 3. 2x2 + x -10 #1 solution 1. x2 + 2x - 15 b = 2 and c = -15 since b is positive and c is negative, the larger factor will be positive and the smaller factor negative the factors of -12 are -1,15 -3, 5 -3 and 5 add to be 2 so our answer is (x - 3)(x + 5) #2 solution 2. 3x3 - 21x2 + 36x Here we have a common factor for all 3terms, 3x , so we must factor it out first. 3x(x2 - 7x + 12) b = -7 and c = 12 since b is negative and c is positive, both factors are negative the factors of 12 are -1,-12 -2,-6 -3, -4 -3 and -4 add to be -7 so our answer is 3x(x - 3)(x - 4) #3 solution 3. 2x2 + x -10 ac = -20 and b = 1 since b is positive and ac is negative, the larger factor is positive and the smaller is negative the factors of -20 are -1,20 -2,10 -4,5 -4 and 5 add to be 1 so we use -4 and 5 to break up the middle term 2x2 - 4x + 5x - 10; note: we have not changed the problem since -4x + 5x =1x Now we can use factoring by grouping (2x2 -4x) + (5x - 10) 2x(x - 2) + 5(x - 2) and our answer is (x -2)(2x + 5)