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

May 20, 2015

$\left({x}^{2} + 5\right) \left({x}^{2} + 2 x - 3\right) = \left({x}^{2} + 5\right) \left(x + 3\right) \left(x - 1\right)$

The factors $\left(x + 3\right)$ and $\left(x - 1\right)$ were found by looking for two numbers $a$ and $b$ such that $a \times b = 3$ and $a - b = 2$.
Then $\left(x + a\right) \left(x - b\right) = {x}^{2} + \left(a - b\right) x - a b$.

This is as far as you can break down the original expression into factors, unless you are allowed complex numbers as coefficients. You can tell that $\left({x}^{2} + 5\right)$ has no smaller factors with real coefficients - which would be linear - because ${x}^{2} + 5 > 0$ for all real values of $x$.

On the other hand, if you are allowed complex coefficients, you can factor as:

$\left({x}^{2} + 5\right) = \left(x + i \sqrt{5}\right) \left(x - i \sqrt{5}\right)$, where $i = \sqrt{- 1}$, so

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

$= \left(x + i \sqrt{5}\right) \left(x - i \sqrt{5}\right) \left(x + 3\right) \left(x - 1\right)$
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