Explain how to factor the following trinomial forms: x²+bx+c and ax²+bx+c. Is there more than one way to factor this? Show your answer using both words and mathematical notation.

David, thanks for your question. The two trinomials you used as examples are very similar : ax

^{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)**