^{2}+bx+c is the general form and x

^{2}+bx+c is the specific case where a = 1.

**when c=0**: ax

^{2}+bx+c --> ax

^{2}+bx

There is a common factor of x in both terms so x(ax+b) = ax

^{2}+bx

Example: 5x

^{2}-4x = x(5x-4) (a=5, b=-4, c=0)

**when b=0 and c<0**: ax

^{2}+bx+c --> ax

^{2}+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: 9x

^{2}-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**: ax

^{2}+bx+c --> x

^{2}+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: x

^{2}-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

one negative.

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)**= x

^{2}-3x-28

**When a=-1**: ax

^{2}+bx+c --> -x

^{2}+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: -x

^{2}+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**ax

^{2}+bx+c: This is more complex than x

^{2}+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: 8x

^{2}+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, 8x

^{2}+2x -21 = (fx+d)(gx+e) =

**(2x-3)(4x+7)**