# Is it possible to factor y= 3x^3+8x^2+3x-2 ? If so, what are the factors?

Feb 12, 2016

$3 {x}^{3} + 8 {x}^{2} + 3 x - 2 = \left(x + 1\right) \left(3 x - 1\right) \left(x + 2\right)$

#### Explanation:

Let $f \left(x\right) = 3 {x}^{3} + 8 {x}^{2} + 3 x - 2$

First notice that $f \left(- 1\right) = - 3 + 8 - 3 - 2 = 0$, so $x = - 1$ is a zero and $\left(x + 1\right)$ is a factor:

$y = 3 {x}^{3} + 8 {x}^{2} + 3 x - 2 = \left(x + 1\right) \left(3 {x}^{2} + 5 x - 2\right)$

We can factor the remaining quadratic by grouping, with the help of an AC Method.

Look for a pair of factors of $A C = 3 \cdot 2 = 6$ which differ by $B = 5$. The pair $6 , 1$ works, so use that to split the middle term before factoring by grouping:

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

$= 3 {x}^{2} + 6 x - x - 2$

$= \left(3 {x}^{2} + 6 x\right) - \left(x + 2\right)$

$= 3 x \left(x + 2\right) - 1 \left(x + 2\right)$

$= \left(3 x - 1\right) \left(x + 2\right)$

Putting it all together:

$3 {x}^{3} + 8 {x}^{2} + 3 x - 2 = \left(x + 1\right) \left(3 x - 1\right) \left(x + 2\right)$