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

Jun 22, 2016

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

#### Explanation:

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

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

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:

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

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

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

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