How do I "solve by factoring" something with an = sign?
Like this: $\displaystyle 3x^2+5x=2$
?
$\displaystyle 3x^2 + 5x = 2$
$\displaystyle 3x^2 + 5x - 2 = 0$
This is something called the "ac method."
Multiply the coefficient of the $\displaystyle x^2$ term by the constant term:
$\displaystyle 3 \cdot (-2) = -6$
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:
$\displaystyle 3x^2 + 5x - 2 = 0$
We want to rewrite the linear (x) term as $\displaystyle 5x = 6x + (-1)x$:
$\displaystyle 3x^2 + (6x - x) - 2 = 0$
Now regroup:
$\displaystyle (3x^2 + 6x) + (-x - 2) = 0$
Factor each of the terms in parenthesis:
$\displaystyle 3x(x + 2) + (-1)(x + 2) = 0$
Now each term has a common factor of $\displaystyle x + 2$, so factor that:
$\displaystyle (3x - 1)(x + 2) = 0$
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
$\displaystyle (3x - 1)(x + 2) = 0$
Thus either
$\displaystyle 3x - 1 = 0 \implies x = \frac{1}{3}$
or
$\displaystyle x + 2 = 0 \implies x = -2$
-Dan