How do you factor #b^2 + 6b -27#?

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Answer 1 · 2015-05-24T09:06:01

Given that the term in #b^2# has coefficient #1# and taking into account the signs of the other coefficients, this may factor as:

#b^2+6b-27 = (b+m)(b-n)#

#= b^2 + (m-n)b - mn#

where #m > 0#, #n > 0#, #m - n = 6# and #mn = 27#.

#27 = 9 xx 3# and #6 = 9 - 3#

so we can choose #m=9#, #n=3# to get:

#b^2+6b-27 = (b+9)(b-3)#