# How do you factor 2b^3+3b^2-8b-12 by grouping?

Feb 9, 2017

$\left(2 b + 3\right) \left(b + 2\right) \left(b - 2\right)$

#### Explanation:

1. If you try to find a connection between 2, 3, 8 and 12. You will find that 2 is the common factor of 2 and 8; 3 is the common factor or 3 and 12.

2.So try group them like above and you will get: $2 b \left({b}^{2} - 4\right) + 3 \left({b}^{2} - 4\right)$

1. And here comes the answer: $\left(2 b + 3\right) \left({b}^{2} - 4\right)$

2. You can further break $\left({b}^{2} - 4\right)$ into $\left(b + 2\right) \left(b - 2\right)$

3. The final result is : $\left(2 b + 3\right) \left(b + 2\right) \left(b - 2\right)$