# Hey, if anybody it would be great and show work thanks. 3x^3+15x^2-12x^2-60x. ?

## factor by grouping

Apr 13, 2017

Assuming the questioner wants solutions to $3 {x}^{3} + 15 {x}^{2} - 12 {x}^{2} - 60 x = 0$, the method is as follows.

#### Explanation:

$3 {x}^{3} + 3 {x}^{2} - 60 x = 0$

$x \left(3 {x}^{3} + 3 x - 60\right) = 0$

$x = 0$

$3 {x}^{2} + 3 x - 60 = 0$

$x = \frac{- 3 \pm \sqrt{9 - 4 \left(3\right) \left(- 60\right)}}{2 \left(3\right)} = \frac{- 3 \pm \sqrt{729}}{6} = \frac{- 3 \pm 27}{6}$

$x = - 5 , 4$

${x}_{1} = 0$

${x}_{2} = 4$

${x}_{3} = - 5$

Apr 13, 2017

The factors are $3 x \left(x + 5\right) \left(x - 4\right)$

#### Explanation:

From the terms in the expression, it is possible that the question is to find the factors of the expression.

We can group the four factors into two pairs:

$\left(3 {x}^{3} + 15 {x}^{2}\right) + \left(- 12 {x}^{2} - 60 x\right) \text{ } \leftarrow$ common factors

=3x^2(x+5)-12x(x+5)"larr common factor and bracket

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