# How do you factor completely: 2x^3 + 14x^2 + 6x + 42?

Jul 17, 2015

$2 \left(x + 7\right) \left({x}^{2} + 3\right)$
Notice that, if you factor out $2 {x}^{2}$ from the first two members and factor out $6$ from the last two members of this equation, you will have the same in both remainders - $\left(x + 7\right)$:
$2 {x}^{3} + 14 {x}^{2} + 6 x + 42 = 2 {x}^{2} \left(x + 7\right) + 6 \left(x + 7\right)$
Now we can use the distributive law and factor out $\left(x + 7\right)$ getting:
$\left(x + 7\right) \left(2 {x}^{2} + 6\right) = 2 \left(x + 7\right) \left({x}^{2} + 3\right)$