How do I "solve by factoring" something with an = sign?

Like this:

?

Printable View

- Aug 31st 2007, 09:28 PMdeathtolife04How do I "solve by factoring" something with an = sign?
How do I "solve by factoring" something with an = sign?

Like this:

? - Aug 31st 2007, 09:37 PMKrizalid
, so

- Aug 31st 2007, 11:51 PMtopsquark

This is something called the "ac method."

Multiply the coefficient of the term by the constant term:

Now write out all the possible pairs of factors of -6:

1, -6

2, -3

3, -2

6, -1

Now add the pairs of factors:

1 + (-6) = -5

2 + (-3) = -1

3 + (-2) = 1

6 + (-1) = 5

*If*your quadratic equation factors over the integers (which is the long term for what they mean by "factoring") then the coefficient of the x term is somewhere in this list. In this case the coefficient of the x term is 5, so we have:

6 + (-1) = 5.

So here's the equation again:

We want to rewrite the linear (x) term as :

Now regroup:

Factor each of the terms in parenthesis:

Now each term has a common factor of , so factor that:

Voila! It is factored. (You can, and should, multiply this back out to check that it matches the original expression.)

From here we will use the property that if the product ab = 0, then either a = 0 or b = 0.

So we know that

Thus either

or

-Dan