How do you factor #4x^2 + 9#?

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Answer 1 · 2017-01-15T21:33:19

There are no simpler factors with Real coefficients, but:

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

Note that #x^2 >= 0# for any Real value of #x#.

Hence #4x^2+9 >= 9# for all Real values of #x#.

So #4x^2+9# has no linear factors with Real coefficients.

It is possible to factor it with Complex coefficients.

The difference of squares identity can be written:

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

In conjunction with #i^2=-1#, we find:

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