Solve z^3-z^2-9z+9

$$z ^ { 3 } - z ^ { 2 } - 9 z + 9$$

Factor

$$\left(z-3\right)\left(z-1\right)\left(z+3\right)$$

Solution Steps

Do the grouping z^{3}-z^{2}-9z+9=\left(z^{3}-z^{2}\right)+\left(-9z+9\right), and factor out z^{2} in the first and -9 in the second group.

$$z^{2}\left(z-1\right)-9\left(z-1\right)$$

Factor out common term z-1 by using distributive property.

$$\left(z-1\right)\left(z^{2}-9\right)$$

Consider z^{2}-9. Rewrite z^{2}-9 as z^{2}-3^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

Rewrite the complete factored expression.

$$\left(z-3\right)\left(z-1\right)\left(z+3\right)$$

Evaluate

$$\left(z-1\right)\left(z^{2}-9\right)$$