# Is it possible to factor y=2x^2 + 9x - 5 ? If so, what are the factors?

Jan 1, 2016

$y = \left(2 x - 1\right) \left(x + 5\right)$

#### Explanation:

As it so happens, any quadratic can be factored into the form

$a {x}^{2} + b x + c$
$= a \left(x - \frac{- b + \sqrt{{b}^{2} - 4 a c}}{2 a}\right) \left(x - \frac{- b - \sqrt{{b}^{2} - 4 a c}}{2 a}\right)$

That big block comes from the quadratic formula, along with the fact that given a polynomial $P \left(x\right)$, if $P \left(a\right) = 0$, then $\left(x - a\right)$ is a factor of $P \left(x\right)$.

However, sometimes (when ${b}^{2} - 4 a c < 0$) the factors involve complex numbers, and there is also a lot to calculate there. Let's see if we can find some real factors without resorting to that right away.

If we have the factorization
$\left(A x + B\right) \left(C x + D\right) = 2 {x}^{2} + 9 x - 5$
then expanding the left hand side and comparing coefficients gives us the system

$\left\{\begin{matrix}A C = 2 \\ A D + B C = 9 \\ B D = - 5\end{matrix}\right.$

If we limit ourselves to integer solutions, then we quickly see from the first and third equations that $A$ and $C$ have to be $\pm 1$ and $\pm 2$ and that $B$ and $D$ have to be $\pm 1$ and ∓$5$. A quick check will show that

$\left\{\begin{matrix}A = 2 \\ B = - 1 \\ C = 1 \\ D = 5\end{matrix}\right.$

is a solution. Thus, substituting back, we have

$\left(2 x - 1\right) \left(x + 5\right) = 2 {x}^{2} + 9 x - 5$