How do you factor completely #9x^2+16 #?

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Answer 1 · 2015-12-02T00:25:07

#9x^2+16# has no simpler factors with Real coefficients, but it does have factors with Complex coefficients:

#9x^2+16 = (3x-4i)(3x+4i)#

If #x in RR# then #x^2 >= 0#, hence #9x^2+16 >= 16 > 0# for all #x in RR#. Hence #9x^2+16# has no linear factors with Real coefficients.

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

If we let #a = 3x# and #b = 4i#, then #a^2=9x^2# and #b^2 = 16i^2 = -16#.

So:

#9x^2+16 = (3x)^2-(4i)^2 = (3x-4i)(3x+4i)#