Solve a^4+1-2a^2

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

Factor

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

Solution Steps

To factor the expression, solve the equation where it equals to 0.

$$a^{4}-2a^{2}+1=0$$

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.

$$a=1$$

By Factor theorem, a-k is a factor of the polynomial for each root k. Divide a^{4}-2a^{2}+1 by a-1 to get a^{3}+a^{2}-a-1. To factor the result, solve the equation where it equals to 0.

$$a^{3}+a^{2}-a-1=0$$

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.

$$a=1$$

By Factor theorem, a-k is a factor of the polynomial for each root k. Divide a^{3}+a^{2}-a-1 by a-1 to get a^{2}+2a+1. To factor the result, solve the equation where it equals to 0.

$$a^{2}+2a+1=0$$

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.

$$a=\frac{-2±\sqrt{2^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$a=\frac{-2±0}{2}$$

Solutions are the same.

$$a=-1$$

Rewrite the factored expression using the obtained roots.

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

Evaluate

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