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

Mar 29, 2015

$\left(3 w + 1\right) \left(w + 9\right) = 0$

Method 1:
When you factor a quadratic expression where the number in front of the squared term is not 1, you are looking for two numbers whose product equals the last term times the coefficient of the squared term and whose sum equals the middle term.

Take the equation: $3 {w}^{2} + 28 w + 9 = 0$

We are looking for two terms whose sum is $28$ and whose product is $27$.

If $a \cdot b = 27$
and $a + b = 28$
$a$ and $b$ should be $1$ and $27$.

So we can split the middle term into two terms:
$3 {w}^{2} + \textcolor{red}{27 w + w} + 9 = 0$

Factor by grouping:
$3 w \left(w + 9\right) + 1 \left(w + 9\right) = 0$
$\left(3 w + 1\right) \left(w + 9\right) = 0$

If it is difficult to see this instinctively, you could always use the quadratic equation.