How do you factor the trinomial #12m^2 + 48m + 96#?

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Answer 1 · 2016-03-31T18:15:39

#12m^2+48m+96 = 12(m+2-2i)(m+2+2i)#

First notice that all of the coefficients are divisible by #12#, so separate that out as a factor first:

#12m^2+48m+96 = 12(m^2+4m+8)#

The remaining quadratic factor is in the form #am^2+bm+c# with #a=1#, #b=4# and #c=8#. This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = 4^2-(4*1*8) = 16 - 32 = -16#

Since this is negative, the quadratic has no simpler factors with Real coefficients.

We can factor it using Complex coefficients, by completing the square as follows:

#m^2+4m+8#

#=m^2+4m+4+4#

#=(m+2)^2+2^2#

#=(m+2)^2-(2i)^2#

#=((m+2)-2i)((m+2)+2i)#

#=(m+2-2i)(m+2+2i)#

Putting it all together:

#12m^2+48m+96 = 12(m+2-2i)(m+2+2i)#