To factor the expression, solve the equation where it equals to $0$.
$$x^{4}-2x^{2}+1=0$$
By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $1$ and $q$ divides the leading coefficient $1$. List all candidates $\frac{p}{q}$.
$$±1$$
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.
$$x=1$$
By Factor theorem, $x-k$ is a factor of the polynomial for each root $k$. Divide $x^{4}-2x^{2}+1$ by $x-1$ to get $x^{3}+x^{2}-x-1$. To factor the result, solve the equation where it equals to $0$.
$$x^{3}+x^{2}-x-1=0$$
By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $-1$ and $q$ divides the leading coefficient $1$. List all candidates $\frac{p}{q}$.
$$±1$$
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.
$$x=1$$
By Factor theorem, $x-k$ is a factor of the polynomial for each root $k$. Divide $x^{3}+x^{2}-x-1$ by $x-1$ to get $x^{2}+2x+1$. To factor the result, solve the equation where it equals to $0$.
$$x^{2}+2x+1=0$$
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 $a$, $2$ for $b$, and $1$ for $c$ in the quadratic formula.
$$x=\frac{-2±\sqrt{2^{2}-4\times 1\times 1}}{2}$$
Do the calculations.
$$x=\frac{-2±0}{2}$$
Solutions are the same.
$$x=-1$$
Rewrite the factored expression using the obtained roots.
