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

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

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Answer 1 · 2016-06-22T21:33:42

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

Explanation:

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

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

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:

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

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

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

$= (\sqrt{3} x - \sqrt{5} i) (\sqrt{3} x + \sqrt{5} i)$ $=(sqrt(3)x-sqrt(5)i)(sqrt(3)x+sqrt(5)i)$

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