when c=0: ax2+bx+c --> ax2+bx
There is a common factor of x in both terms so x(ax+b) = ax2+bx
Example: 5x2-4x = x(5x-4) (a=5, b=-4, c=0)
when b=0 and c<0: ax2+bx+c --> ax2+c
let d = -c. d>0 because c<0 and ax2+c = ax2-d
The difference of two squares --> (x√a + √d)(x√a - √d) = ax2-d
d = -c, so (x√a + √(-c))(x√a - √(-c)) = ax2+c
Remember that this only works when b=0 and c<0 and can only be factored if a is a
perfect square and if -c is a perfect square.
Example: 9x2-25 = (x√9 + √(25))(x√9 - √(25)) = (3x + 5)(3x - 5)
or (-3x + 5)(-3x - 5)
(a=9, b=0, c=-25)
when a=1: ax2+bx+c --> x2+bx+c
There are a number of ways to approach factoring a trinomial in this form, but
no matter the technique you need a result which looks like (x + d)(x + e)
You need to find a d and e such that (d)(e) = c AND dx + ex = bx, or d + e = b
Not every trinomial can be factored!
Example: x2-3x-28 (a=1, b=-3, c=-28)
need to find d and e such that (d)(e) = -28 and d + e = -3
Since c is negative, then d and e will be a pair with one positive and
The factors of 28 are 1&28, 2&14, 4&7. For -28 one of the pair is negative.
We need to use one of these pairs as d and e with one of them negative.
Using d=4 and e=-7, (4)(-7) = -28 and 4+(-7)=-3
(x + d)(x + e) = (x + 4)(x - 7) = x2-3x-28
When a=-1: ax2+bx+c --> -x2+bx+c
Same idea, but your result will look like (x + d)(-x + e)
You need to find a d and e such that (d)(e) = c AND ex - dx = bx, or e-d = b
Example: -x2+2x+15 (a=-1, b=2, c=15)
The factors of 15 are 1&15, 3&5, -1&-15, -3&-5.
We need to use one of these pairs as d and e.
Use d=3 and e=5, (3)(5)=15, 5-3 = 2 --> (x+3)(-x+5)
OR use d=-5 and e=-3, (-5)(-3)=15, -3 -(-5) = 2 --> (x-5)(-x-3)
The general case ax2+bx+c: This is more complex than x2+bx+c because you must now find two more
constants, f and g so that your result will be (fx + d)(gx +e)
If you expand this form you get (fg)x2 + (dg +ef)x +de
This requires that (d)(e) = c AND (f)(g) = a, AND dg +ef = b
Example: 8x2+2x -21 (a=8, b=2, c=-21)
Factors of 21 are 1&21, 3&7. -21 means one + one -
Factors of 8 are 1&8, 2&4. Need both + or both -
Since b is small, experience tells me to try combinations of 3&7, and 2&4
Try d=3, e=-7, f=2, g=4.
3(-7)=-21, 2(4) = 8, (3)(4)+(-7)(2)=-2 (needs to =2) DOESN'T WORK
Try swapping signs for d and e, d=-3, e=7, f=2, g=4
-3(7)=-21, 2(4)=8, -3(4)+7(2)=2 WORKS!
So, 8x2+2x -21 = (fx+d)(gx+e) = (2x-3)(4x+7)