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)#