# How do you factor 6x^3+10x^2-32?

Sep 30, 2015

$2 \left(x - \frac{4}{3}\right) \left(3 {x}^{2} + 8 x + 12\right)$
After taking out 2 as a common factor it becomes $2 \left(3 {x}^{3} + 5 {x}^{2} - 16\right)$
In this 3rd degree polynomial, the product of its roots (zeros) would be $- \frac{16}{3}$. Try $x = \frac{4}{3}$, it is found that it is on of the roots. Hence $\left(x - \frac{4}{3}\right)$ would be a factor. Now carry out long or synthetic division of the polynomial by $\left(x - \frac{4}{3}\right)$ to have $3 {x}^{2} + 8 x + 12$ as a quotient.
The required factorisation would be thus $2 \left(x - \frac{4}{3}\right) \left(3 {x}^{2} + 8 x + 12\right)$