How do you factor completely: #b^2 -7b -44#?

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How do you factor completely: #b^2 -7b -44#?

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Answer 1 · 2015-07-18T03:00:56

I found:
#(b-11)(b+4)#

Explanation:

I would first try to solve the equation:
#b^2-7b-44=0# using the Quadratic Formula:
#b_(1,2)=(7+-sqrt(49+176))/2=(7+-15)/2=#
#b_1=(7+15)/2=11#
#b_2=(7-15)/2=-4#

So we can use these two results (with opposite signs) and get:
#(b-11)(b+4)=b^2-7b-44#

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How do you factor completely: #b^2 -7b -44#?

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