Solve k^4-8k^2-9

$$k ^ { 4 } - 8 k ^ { 2 } - 9$$

Factor

$$\left(k-3\right)\left(k+3\right)\left(k^{2}+1\right)$$

Solution Steps

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

$$k^{4}-8k^{2}-9=0$$

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.

$$k=3$$

By Factor theorem, k-p is a factor of the polynomial for each root p. Divide k^{4}-8k^{2}-9 by k-3 to get k^{3}+3k^{2}+k+3. To factor the result, solve the equation where it equals to 0.

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 3 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±3,±1$$

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.

$$k=-3$$

By Factor theorem, k-p is a factor of the polynomial for each root p. Divide k^{3}+3k^{2}+k+3 by k+3 to get k^{2}+1. To factor the result, solve the equation where it equals to 0.

$$k^{2}+1=0$$

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 1 for c in the quadratic formula.

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

Do the calculations.

$$k=\frac{0±\sqrt{-4}}{2}$$

Polynomial k^{2}+1 is not factored since it does not have any rational roots.

$$k^{2}+1$$

Rewrite the factored expression using the obtained roots.

$$\left(k-3\right)\left(k+3\right)\left(k^{2}+1\right)$$

Evaluate

$$k^{4}-8k^{2}-9$$