Solve x^4-2x^2+1

$$x ^ { 4 } - 2 x ^ { 2 } + 1$$

Factor

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

Solution Steps

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.

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

Evaluate

$$\left(x^{2}-1\right)^{2}$$