Solve p^4-2p^2-8

$$p ^ { 4 } - 2 p ^ { 2 } - 8$$

Factor

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

Solution Steps

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

$$p^{4}-2p^{2}-8=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -8 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±8,±4,±2,±1$$

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.

$$p=2$$

By Factor theorem, p-k is a factor of the polynomial for each root k. Divide p^{4}-2p^{2}-8 by p-2 to get p^{3}+2p^{2}+2p+4. To factor the result, solve the equation where it equals to 0.

$$p^{3}+2p^{2}+2p+4=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 4 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±4,±2,±1$$

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.

$$p=-2$$

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

$$p^{2}+2=0$$

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 2 for c in the quadratic formula.

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

Do the calculations.

$$p=\frac{0±\sqrt{-8}}{2}$$

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

$$p^{2}+2$$

Rewrite the factored expression using the obtained roots.

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

Evaluate

$$p^{4}-2p^{2}-8$$