Factoring Quadratic Trinomials via AC Test?

Question

Okay, so we finished discussing the Quadratic Trinomials. We were taught how to solve it via trial and error. Now, they told us that the AC Test was a shorter way and would spend less time than the trial and error. Can anyone explain? I didn't really get it when I searched the internet for it...

Answer 1

http://www.jamesbrennan.org/algebra/polynomials/summary_of_steps_to_factor_quadr.htm

Factoring by Grouping

Given a general quadratic ax^2 + bx + c = 0,

1. Find the product ac
2. Find two factors of ac that add to give b
3. Split the middle term into two terms, using the numbers found in step 2.
4. Group the four terms into two pairs
5. Factor out the common binomial

Example: 3x^2 - 2x - 1

1. Find the product ac.
Since a = 3 and c = -1, then ac = -3

2. Find two factors of ac that add to b.
b = -2, ac = -3, two factors of ac that add to b are
(-3) and (+1)

3. Split the middle term into two terms, using the numbers found in step 2.

3x^2 - 2x - 1 = 3x^2 - 3x + x - 1

4. Group the four terms into two pairs

3x(x - 1) + (x - 1)

5. Factor out the common binomial.

(x - 1) (3x + 1)

There we go!

Let's try another example:
Example 2:

10x^2 - 7x - 12

ac = -120
Two factors of -120 that add to -7 are -15 and +8.
Splitting -7x into -15x and +8x, we get

10x^2 - 15x + 8x - 12

Grouping, we get

5x(2x - 3) + 4(2x - 3)

And now we group normally.

(2x - 3)(5x + 4).

Wow... it really is faster. (This is actually the first time I've learned this).

Answer 2

I believe you are looking for the AC Method. Heres a link to help you understand it.


http://www.richland.edu/james/misc/acmeth.html

I wish I knew the name of the method I use, I would refer you too it.