Solve 2x^4-x^2-1

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

Factor

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

Solution Steps

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

$$2x^{4}-x^{2}-1=0$$

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 2. List all candidates \frac{p}{q}.

$$±\frac{1}{2},±1$$

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 2x^{4}-x^{2}-1 by x-1 to get 2x^{3}+2x^{2}+x+1. To factor the result, solve the equation where it equals to 0.

$$2x^{3}+2x^{2}+x+1=0$$

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 2. List all candidates \frac{p}{q}.

$$±\frac{1}{2},±1$$

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 2x^{3}+2x^{2}+x+1 by x+1 to get 2x^{2}+1. To factor the result, solve the equation where it equals to 0.

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

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 2 for a, 0 for b, and 1 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 2\times 1}}{2\times 2}$$

Do the calculations.

$$x=\frac{0±\sqrt{-8}}{4}$$

Polynomial 2x^{2}+1 is not factored since it does not have any rational roots.

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

Rewrite the factored expression using the obtained roots.

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

Evaluate

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