Solve x^4+x^2-20

$$x ^ { 4 } + x ^ { 2 } - 20$$

Factor

$$\left(x-2\right)\left(x+2\right)\left(x^{2}+5\right)$$

Solution Steps

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

$$x^{4}+x^{2}-20=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -20 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±20,±10,±5,±4,±2,±1$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}+x^{2}-20 by x-2 to get x^{3}+2x^{2}+5x+10. To factor the result, solve the equation where it equals to 0.

$$x^{3}+2x^{2}+5x+10=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 10 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±10,±5,±2,±1$$

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.

$$x=-2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+2x^{2}+5x+10 by x+2 to get x^{2}+5. To factor the result, solve the equation where it equals to 0.

$$x^{2}+5=0$$

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 5 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\times 5}}{2}$$

Do the calculations.

$$x=\frac{0±\sqrt{-20}}{2}$$

Polynomial x^{2}+5 is not factored since it does not have any rational roots.

$$x^{2}+5$$

Rewrite the factored expression using the obtained roots.

$$\left(x-2\right)\left(x+2\right)\left(x^{2}+5\right)$$

Evaluate

$$x^{4}+x^{2}-20$$