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:

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

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