Factoring
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)
1. x2 + 2x - 15: solution
2. 3x3 - 21x2 + 36x :solution
3. 2x2 + x -10: 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. 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. 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)