# How do you simplify (15n^2 – 5n + 45) ÷ 5?

$\left(15 {n}^{2} - 5 n + 45\right) \div 5 = 3 {n}^{2} - n + 9$
$3 {n}^{2} - n + 9$ is of the form $a {n}^{2} + b n + c$, with $a = 3$, $b = - 1$ and $c = 9$.
${b}^{2} - 4 a c = {\left(- 1\right)}^{2} - 4 \cdot 3 \cdot 9 = 1 - 108 = - 107$
which is negative, so there are no real values of $n$ satisfying $3 {n}^{2} - n + 9 = 0$ and no linear factors with real coefficients.