Solve -x^4+6x^2+27

$$- { x }^{ 4 } +6 { x }^{ 2 } +27$$

Factor

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

Solution Steps

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

$$-x^{4}+6x^{2}+27=0$$

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^{4}+6x^{2}+27 by x-3 to get -x^{3}-3x^{2}-3x-9. To factor the result, solve the equation where it equals to 0.

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -9 and q divides the leading coefficient -1. List all candidates \frac{p}{q}.

$$±9,±3,±1$$

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}-3x-9 by x+3 to get -x^{2}-3. To factor the result, solve the equation where it equals to 0.

$$-x^{2}-3=0$$

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 -3 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\left(-1\right)\left(-3\right)}}{-2}$$

Do the calculations.

$$x=\frac{0±\sqrt{-12}}{-2}$$

Polynomial -x^{2}-3 is not factored since it does not have any rational roots.

$$-x^{2}-3$$

Rewrite the factored expression using the obtained roots.

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

Evaluate

$$27+6x^{2}-x^{4}$$