# How do you solve 3w^2 + 28w + 9 = 0 by factoring?

Mar 27, 2015

The solutions are: $- \frac{1}{3}$, $- 9$

$3 {w}^{2} + 28 w + 9 = 0$

If it can be factored using whole numbers it must be

$\left(3 w + {\text{something"_1)(w+"something}}_{2}\right)$
So that the ${w}^{2}$ term has a coefficient of $3$

${\text{something"_1 xx "something}}_{2}$ must be $9$
(the product of the constants (the lasts) must be the constant $9$)
To get a product of $9$ using whole numbers our choices are
$1 \cdot 9$ or $3 \cdot 3$.

But the first terms are different so we need to consider the choices:
$\left(3 w + 1\right) \left(w + 9\right)$
$\left(3 w + 9\right) \left(w + 1\right)$

$\left(3 w + 3\right) \left(w + 3\right)$ (Notice this would make the whole thing divisible by $3$ -- which it isn't. This won't work.)

A quick check will convince you that

$\left(3 w + 1\right) \left(w + 9\right) = 3 {w}^{2} + 28 w + 9$

So, solving $3 {w}^{2} + 28 w + 9 = 0$ is the same as solving:
$\left(3 w + 1\right) \left(w + 9\right) = 0$

And the only way for a product of numbers to be $0$ is for at least one of the numbers to be $0$. So we need:

$3 w + 1 = 0$ or $w + 9 = 0$

In the first case we get $w = - \frac{1}{3}$, in the second $w = - 9$.

The solutions are: $- \frac{1}{3}$, $- 9$