# How do you factor completely -4y^3+3y^2+8y-6?

Apr 6, 2018

Correct answer is $- \left(4 y - 3\right) \left({y}^{2} - 2\right)$

#### Explanation:

So you start off with the first two terms which are $- 4 {y}^{3}$ and $3 {y}^{2}$. Now you must think what numbers could we possibly have that will multiply to them.

Only the factors $4 , 1 , 3$ and $1$ work but we cannot ignore the negative. If we were to place the negative in the other set of ( ), the answer would not come out right.

So for now, it should be $- \left(4 y - 3\right) \left({y}^{2}\right)$

So what number will give us $- 6$? Since $- 3$ times something gives us $- 6$, it has to be $- 2$ because once we do the FOIL method and multiply it by the negative, we are not going to get $- 6$ because if it were to be a positive two, it would still give us a negative $6$ but since the negative is being multiplied to it, we would receive $- 6$.

So our final answer has to be: $- \left(4 y - 3\right) \left({y}^{2} - 2\right)$