Solve x^4-18x^2+81

$${ x }^{ 4 } -18 { x }^{ 2 } +81$$

Factor

$$\left(x-3\right)^{2}\left(x+3\right)^{2}$$

Solution Steps

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

$$x^{4}-18x^{2}+81=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 81 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±81,±27,±9,±3,±1$$

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.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}-18x^{2}+81 by x-3 to get x^{3}+3x^{2}-9x-27. To factor the result, solve the equation where it equals to 0.

$$x^{3}+3x^{2}-9x-27=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -27 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±27,±9,±3,±1$$

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.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+3x^{2}-9x-27 by x-3 to get x^{2}+6x+9. To factor the result, solve the equation where it equals to 0.

$$x^{2}+6x+9=0$$

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, 6 for b, and 9 for c in the quadratic formula.

$$x=\frac{-6±\sqrt{6^{2}-4\times 1\times 9}}{2}$$

Do the calculations.

$$x=\frac{-6±0}{2}$$

Solutions are the same.

$$x=-3$$

Rewrite the factored expression using the obtained roots.

$$\left(x-3\right)^{2}\left(x+3\right)^{2}$$

Evaluate

$$\left(x^{2}-9\right)^{2}$$