# How do you Factor completely 2x^3+5x^2-37x-60?

Jul 15, 2018

$\left(x - 4\right) \left(x + 5\right) \left(2 x + 3\right)$

#### Explanation:

Note that $x = 4$ is a solution
$2 \cdot {4}^{3} + 5 \cdot {4}^{2} - 37 \cdot 4 - 60 = 128 + 80 - 148 - 60 = 0$
and

$x = - 5$

$- 2 \cdot {5}^{3} + 5 \cdot 25 + 37 \cdot 5 - 60 = - 250 + 125 + 185 - 60 = 0$
so we can

$2 \cdot {x}^{3} + 5 \cdot {x}^{2} - 37 \cdot x - 60$
divide by $\left(x - 4\right) \left(x + 5\right)$
and we get $2 x + 3$
so the completely factorization is given by

$\left(x - 4\right) \left(x + 5\right) \left(2 x + 3\right)$

Jul 15, 2018

$\left(x - 4\right) \left(x + 5\right) \left(2 x + 3\right)$

#### Explanation:

$2 {x}^{3} + 5 {x}^{2} - 37 x - 60$

=$2 {x}^{3} - 8 {x}^{2} + 13 {x}^{2} - 52 x + 15 x - 60$

=$2 {x}^{2} \cdot \left(x - 4\right) + 13 x \cdot \left(x - 4\right) + 15 \cdot \left(x - 4\right)$

=$\left(x - 4\right) \cdot \left(2 {x}^{2} + 13 x + 15\right)$

=$\left(x - 4\right) \cdot \left(2 {x}^{2} + 10 x + 3 x + 15\right)$

=(x-4)((2x*(x+5)+3*(x+5))

=$\left(x - 4\right) \left(x + 5\right) \left(2 x + 3\right)$