More:

__Factoring Quadratic Trinomials__

(when x^{2} coefficient is not 1)

Now we know what to do if a = 1, in ax^{2} + bx + c...

But what if it's not? What if it's 2? Or 3? Or -1? (*fake screaming*)

It's just a process that's a little more complicated, and
sometimes called **factoring by grouping** (but I wouldn't take my word for
it - I've heard factoring by grouping can mean all factoring polynomials).

First, make sure that you really know what a, b, and c stand for. Then read this:

2x^{2} + 12x + 16

1. Find product ac.2 * 16 = 322. Find two factors of ac that add up to b
(call them y and z).4 and 83. Rewrite the expression so it is ax^{2}
+ yx + zx + c2x ^{2} + 4x + 8x + 164. Put parentheses around the first two
terms, and the last two.(2x ^{2} + 4x) + (8x + 16)5. Factor the new polynomials (two terms
inside the each of the sets of parentheses).2x(x + 2) + 8(x + 2) 6. Now, the polynomials in the parentheses
should be the same. Therefore, you can factor them out of the
others.(2x + 8)(x + 2) |

So, ready for another one?

**3x ^{2} - 8x + 5**

ac = 15

-3*-5 = 15, -3 + -5 = -8

3x

(3x

3x(x - 1) + -5(x - 1) **Note: You

**-x ^{2} + 3x + 4**

ac = -4

-1*4 = -4, -1 + 4 = 3

(-x

-x(x + 1) + 4(x + 1)

Click here
to go back to factoring quadratic trinomials.

Click here to go back
to the index.