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:

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

How do you factor the trinomial 12m^2 + 48m + 9612m^2 + 48m + 96?