# How do you factor n^2-9n+20?

May 28, 2015

${n}^{2} - 9 n + 20 = \left(n - 4\right) \left(n - 5\right)$

Here you can use the quadratic formula:
let $p \left(n\right) = a {n}^{2} + b n + c$
$\Delta = {b}^{2} - 4 a c$
${n}_{1 , 2} = \frac{- b \pm \sqrt{\Delta}}{2 a}$
Then $p \left(n\right) = a \left(n - {n}_{1}\right) \left(n - {n}_{2}\right)$.

In our case we have $a = 1 , b = - 9 , c = 20$ so
$\Delta = 1 , {n}_{1} = 4 , {n}_{2} = 5$