How do you factor 8b^2-59b+21?

May 13, 2016

$8 {b}^{2} - 59 b + 21 = \left(8 b - 3\right) \left(b - 7\right)$

Explanation:

Use an AC method:

Find a pair of factors of $A C = 8 \cdot 21 = 168$ with sum $B = 59$.

Note that since $59$ is odd, one of the factors must be odd too. So the other factor is divisible by $8$, and we soon find that the pair $56 , 3$ works.

Use this pair to split the middle term and factor by grouping:

$8 {b}^{2} - 59 b + 21$

$= 8 {b}^{2} - 56 b - 3 b + 21$

$= \left(8 {b}^{2} - 56 b\right) - \left(3 b - 21\right)$

$= 8 b \left(b - 7\right) - 3 \left(b - 7\right)$

$= \left(8 b - 3\right) \left(b - 7\right)$