To factor the expression, solve the equation where it equals to $0$.
$$x^{4}-22x^{2}-75=0$$
By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $-75$ and $q$ divides the leading coefficient $1$. List all candidates $\frac{p}{q}$.
$$±75,±25,±15,±5,±3,±1$$
Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.
$$x=5$$
By Factor theorem, $x-k$ is a factor of the polynomial for each root $k$. Divide $x^{4}-22x^{2}-75$ by $x-5$ to get $x^{3}+5x^{2}+3x+15$. To factor the result, solve the equation where it equals to $0$.
$$x^{3}+5x^{2}+3x+15=0$$
By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $15$ and $q$ divides the leading coefficient $1$. List all candidates $\frac{p}{q}$.
$$±15,±5,±3,±1$$
Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.
$$x=-5$$
By Factor theorem, $x-k$ is a factor of the polynomial for each root $k$. Divide $x^{3}+5x^{2}+3x+15$ by $x+5$ to get $x^{2}+3$. To factor the result, solve the equation where it equals to $0$.
$$x^{2}+3=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. Substitute $1$ for $a$, $0$ for $b$, and $3$ for $c$ in the quadratic formula.
$$x=\frac{0±\sqrt{0^{2}-4\times 1\times 3}}{2}$$
Do the calculations.
$$x=\frac{0±\sqrt{-12}}{2}$$
Polynomial $x^{2}+3$ is not factored since it does not have any rational roots.
$$x^{2}+3$$
Rewrite the factored expression using the obtained roots.
