# How do you factor 8x^3+28x^2+24x ?

May 10, 2015

$8 {x}^{3} + 28 {x}^{2} + 24 x = 4 x \left(2 {x}^{2} + 7 x + 6\right) = 4 x \left(2 x + 3\right) \left(x + 2\right)$.

To find this, first note that all the terms are divisible by $4$ and by $x$. Dividing through by $4 x$ yields the quadratic $2 {x}^{2} + 7 x + 6$.

If this has linear factors with integer coefficients they must be of the form $\left(2 x + a\right)$ and $\left(x + b\right)$ in order that the product starts with $2 {x}^{2}$, where $a b = 6$ and $2 b + a = 7$. From this, it's easy to see that $a = 3$ and $b = 2$, so $2 {x}^{2} + 7 x + 6 = \left(2 x + 3\right) \left(x + 2\right)$.