How do you factor 6b^2 + b - 2?

May 31, 2018

$\left(2 b - 1\right) \left(3 b + 2\right)$

Explanation:

$\text{using the a-c method to factor the quadratic}$

$\text{the factors of the product } 6 \times - 2 = - 12$

$\text{which sum to + 1 are - 3 and + 4}$

$\text{split the middle term using these factors}$

$6 {b}^{2} - 3 b + 4 b - 2 \leftarrow \textcolor{b l u e}{\text{factor by grouping}}$

$= \textcolor{red}{3 b} \left(2 b - 1\right) \textcolor{red}{+ 2} \left(2 b - 1\right)$

$\text{take out the "color(blue)"common factor } \left(2 b - 1\right)$

$= \left(2 b - 1\right) \left(\textcolor{red}{3 b + 2}\right)$

$6 {b}^{2} + b - 2 = \left(2 b - 1\right) \left(3 b + 2\right)$