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

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

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Answer 1 · 2016-05-13T18:45:12

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

Explanation:

Use an AC method:

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

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

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

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

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

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

$= 8 b (b - 7) - 3 (b - 7)$ $=8b(b-7)-3(b-7)$

$= (8 b - 3) (b - 7)$ $=(8b-3)(b-7)$

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