Solve a^4-6a^2-27

$$a ^ { 4 } - 6 a ^ { 2 } - 27$$

Factor

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

Solution Steps

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

$$a^{4}-6a^{2}-27=0$$

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.

$$a=3$$

By Factor theorem, a-k is a factor of the polynomial for each root k. Divide a^{4}-6a^{2}-27 by a-3 to get a^{3}+3a^{2}+3a+9. To factor the result, solve the equation where it equals to 0.

$$a^{3}+3a^{2}+3a+9=0$$

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.

$$a=-3$$

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

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

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.

$$a=\frac{0±\sqrt{0^{2}-4\times 1\times 3}}{2}$$

Do the calculations.

$$a=\frac{0±\sqrt{-12}}{2}$$

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

$$a^{2}+3$$

Rewrite the factored expression using the obtained roots.

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

Evaluate

$$a^{4}-6a^{2}-27$$