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

1 Answer
George C.
May 13, 2016

#8b^2-59b+21=(8b-3)(b-7)#

Explanation:

Use an AC method:

Find a pair of factors of #AC = 8*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:

#8b^2-59b+21#

#=8b^2-56b-3b+21#

#=(8b^2-56b)-(3b-21)#

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

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

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