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

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

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

Like this: $3x^2+5x=2$

?

2. $3x^2+(6x-x)-2=0\iff(3x^2+6x)-(x+2)=0$, so $(x+2)(3x-1)=0$

3. Originally Posted by deathtolife04
How do I "solve by factoring" something with an = sign?

Like this: $3x^2+5x=2$

?
$3x^2 + 5x = 2$

$3x^2 + 5x - 2 = 0$

This is something called the "ac method."

Multiply the coefficient of the $x^2$ term by the constant term:
$3 \cdot (-2) = -6$

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:
$3x^2 + 5x - 2 = 0$

We want to rewrite the linear (x) term as $5x = 6x + (-1)x$:
$3x^2 + (6x - x) - 2 = 0$

Now regroup:
$(3x^2 + 6x) + (-x - 2) = 0$

Factor each of the terms in parenthesis:
$3x(x + 2) + (-1)(x + 2) = 0$

Now each term has a common factor of $x + 2$, so factor that:
$(3x - 1)(x + 2) = 0$

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
$(3x - 1)(x + 2) = 0$

Thus either
$3x - 1 = 0 \implies x = \frac{1}{3}$
or
$x + 2 = 0 \implies x = -2$

-Dan