How do you factor $7ab^2 - 77ab + 168a$ ?

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Answer 1 · 2018-04-09T07:58:32

$7ab^2-77ab+168a = 7a(b-8)(b-3)$

Given:

$7ab^2-77ab+168a$

Note that all of the terms are divisible by $7a$ , so we can separate that out as a factor first to find:

$7ab^2-77ab+168a = 7a(b^2-11b+24)$

To factor the remaining quadratic we can find a pair of factors of $24$ with sum $11$ . The pair $8, 3$ works, so we find:

$b^2-11b+24 = (b-8)(b-3)$

So:

$7ab^2-77ab+168a = 7a(b-8)(b-3)$

How do you factor 7ab^2 - 77ab + 168a7ab^2 - 77ab + 168a?