# Is it possible to factor y=x^2-x-56 ? If so, what are the factors?

Mar 5, 2016

#### Answer:

$\left(x + 7\right) \left(x - 8\right)$

#### Explanation:

For finding whether or not $a {x}^{2} + b x + c = 0$ can be factorized or not, we find the value ${b}^{2} - 4 a c$.

As the equation is x^2−x−56, ${b}^{2} - 4 a c = {1}^{2} - 4 \times 1 \times \left(- 56\right) = 1 + 224 = 225$. As it is a perfect square, it should be possible to factorize it.

Hence for factorizing, we split $a c$ (the product of coefficient of ${x}^{2}$ and independent term into two factors, whose sum is $b$, the coefficient of $x$.

In the polynomial x^2−x−56, the product is $- 56$ and hence factors whose sum is $- 1$, these would be $- 8$ and $7$. Hence splitting middle term this way, we get

x^2−x−56=x^2−8x+7x−56 i.e.

$x \left(x - 8\right) + 7 \left(x - 8\right)$ or

$\left(x + 7\right) \left(x - 8\right)$