# How do you factor completely x^4 + 7x^2 + 10?

Jan 3, 2016

You need to change the variable in order to find the roots.

#### Explanation:

Let's say $P \left(x\right) = {x}^{4} + 7 {x}^{2} + 10$ and let's say $X = {x}^{2}$. So we're now studying $P \left(X\right) = {X}^{2} + 7 X + 10$, which is a classic trinomial we can solve easily.

In order to use the quadratic formula, we first need to calculate $\Delta = {b}^{2} - 4 a c$. Here, $\Delta = {7}^{2} - 4 \cdot 10 = 9$. So this polynomial has 2 real roots.

By the quadratic formula, the roots are given by $\frac{- b \pm \sqrt{\Delta}}{2} a$.

${X}_{1} = \frac{- 7 - 3}{2} = - \frac{10}{2} = - 5$ and ${X}_{2} = \frac{- 7 + 3}{2} = - \frac{4}{2} = - 2$.

So $P \left(X\right) = \left(X + 5\right) \left(X + 2\right)$. We now switch back to the original variable, hence $P \left(x\right) = \left({x}^{2} + 5\right) \left({x}^{2} + 2\right)$. If we want to factorise it completely, there is still some work to do with complex numbers.

I will only show you how to solve ${x}^{2} + 5 = 0$ in $\mathbb{C}$ because it's the exact same way to solve ${x}^{2} + 2 = 0$.

In $\mathbb{C}$, ${x}^{2} + 5 = 0 \iff {x}^{2} = - 5 \iff x = \pm i \sqrt{5}$. So the complete factorization of $P$ in $\mathbb{C}$ is $\left(x - i \sqrt{5}\right) \left(x + i \sqrt{5}\right) \left(x - i \sqrt{2}\right) \left(x + i \sqrt{2}\right)$.