How do you factor $3x^2 + 5$ ?

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Answer 1 · 2016-06-22T21:33:42

$3x^2+5=(sqrt(3)x-sqrt(5)i)(sqrt(3)x+sqrt(5)i)$

Note that if $x$ is a Real number then $x^2 >= 0$ .

Hence $3x^2+5 > 0$ .

So this quadratic has no Real zeros and no linear factors with Real coefficients.

It can be factored with Complex numbers.

Use the difference of squares identity:

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

with $a=sqrt(3)x$ and $b=sqrt(5)i$ as follows:

$3x^2+5=(sqrt(3)x)^2-(sqrt(5)i)^2$

$=(sqrt(3)x-sqrt(5)i)(sqrt(3)x+sqrt(5)i)$

How do you factor 3x^2 + 53x^2 + 5?