Solve z^4-13z^2+36

$$z ^ { 4 } - 13 z ^ { 2 } + 36$$

Factor

$$\left(z-3\right)\left(z-2\right)\left(z+2\right)\left(z+3\right)$$

Solution Steps

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

$$z^{4}-13z^{2}+36=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 36 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±36,±18,±12,±9,±6,±4,±3,±2,±1$$

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.

$$z=2$$

By Factor theorem, z-k is a factor of the polynomial for each root k. Divide z^{4}-13z^{2}+36 by z-2 to get z^{3}+2z^{2}-9z-18. To factor the result, solve the equation where it equals to 0.

$$z^{3}+2z^{2}-9z-18=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -18 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±18,±9,±6,±3,±2,±1$$

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.

$$z=-2$$

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

$$z^{2}-9=0$$

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, 0 for b, and -9 for c in the quadratic formula.

$$z=\frac{0±\sqrt{0^{2}-4\times 1\left(-9\right)}}{2}$$

Do the calculations.

$$z=\frac{0±6}{2}$$

Solve the equation z^{2}-9=0 when ± is plus and when ± is minus.

$$z=-3 z=3$$

Rewrite the factored expression using the obtained roots.

$$\left(z-3\right)\left(z-2\right)\left(z+2\right)\left(z+3\right)$$

Evaluate

$$z^{4}-13z^{2}+36$$