# Can y= x^2+7x-30  be factored? If so what are the factors ?

Jan 4, 2016

${x}^{2} + 7 x - 30 = \left(x + 10\right) \left(x - 3\right)$

#### Explanation:

Find a pair of factors of $30$ which multiply to give $30$ and differ by $7$. The pair $10$, $3$ works, hence:

${x}^{2} + 7 x - 30 = \left(x + 10\right) \left(x - 3\right)$

${x}^{2} + 7 x - 30$ is in the form $a {x}^{2} + b x + c$ with $a = 1$, $b = 7$ and $c = - 30$.

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {7}^{2} - \left(4 \times 1 \times - 30\right) = 49 + 120 = 169 = {13}^{2}$

Since this is a perfect square, the quadratic has two linear factors with rational coefficients.

Rather than search for a suitable pair of factors of $30$ we could use the quadratic formula to find the zeros of ${x}^{2} + 7 x - 30$ and hence its factors:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{- b \pm \sqrt{\Delta}}{2 a}$

$= \frac{- 7 \pm 13}{2}$

That is $x = - 10$ or $x = 3$

Hence factors $\left(x + 10\right)$ and $\left(x - 3\right)$