How do you factor 4v^2-9v-9?

Apr 14, 2015

$4 {v}^{2} - 9 v - 9$

$= \left(4 v - 3\right) \left(v + 3\right)$

But how to get there is a little more complex.

Start by looking at factor pairs of $4$
(that is values $\left(a , b\right)$ such that $a \times b = 4$)
and
factor pairs of $9$
(again, values $\left(p , q\right)$ such that $p \times q = 9$)

There are only a very few possibilities
$4 = \left(4 \times 1\right)$ or $\left(2 \times 2\right)$ or negative versions of these pairs
$- 9 = \left(9 \times - 1\right)$ or $\left(3 , - 3\right)$ or negative versions of these pairs

We are looking for values $\left(a , b\right)$ and $\left(p , q\right)$
such that
$\left(a v + p\right) \left(b v + q\right) = 4 {v}^{2} - 9 v - 9$

Because of our restrictions on possible factor pair values
we simply need to find the two pair factors that give
$a q + b p = - 9$

Since there are not a lot of combinations
the two pairs $\left(4 , 1\right)$ and $\left(- 3 , 3\right)$ soon become obvious.