# How do you factor the expression 25x^2 - 35x - 30?

Dec 2, 2015

$25 {x}^{2} - 35 x - 30 = 5 \left(x - 2\right) \left(5 x + 3\right)$

#### Explanation:

First, we can easily see that each term has a factor of $5$, so let's take that out

$25 {x}^{2} - 35 x - 30 = 5 \left(5 {x}^{2} - 7 x - 6\right)$

Now we just need to factor $5 {x}^{2} - 7 x - 6$.

We could use the quadratic formula to find the roots of the expression, however let's see if we can find an integer solution without resorting to that.

We are looking for integers $A , B , C , D$ such that

$5 {x}^{2} - 7 x - 6 = \left(A x + B\right) \left(C x + D\right)$

$\implies 5 {x}^{2} - 7 x - 6 = A C {x}^{2} + \left(A D + B C\right) x + B D$

$\implies \left\{\begin{matrix}A C = 5 \\ A D + B C = - 7 \\ B D = - 6\end{matrix}\right.$

From the first equation, we know that $A$ and $D$ are $1$ and $5$.

From the third equation, we know that $B$ and $C$ are one of
$1$ and $- 6$
$- 1$ and $6$
$2$ and $- 3$
$- 2$ and $3$

Testing out which set of pairs fulfills the second equation lets us find a solution

$A = 1$
$B = - 2$
$C = 5$
$D = 3$

Then substituting those back in gives us

$5 {x}^{2} - 7 x - 6 = \left(x - 2\right) \left(5 x + 3\right)$

But we mustn't forget the original $5$ we factored out, and so we have the final result of

$25 {x}^{2} - 35 x - 30 = 5 \left(x - 2\right) \left(5 x + 3\right)$