Solve rho(z)=z^4-9z^2-z^3a+9za

$$\rho ( z ) = z ^ { 4 } - 9 z ^ { 2 } - z ^ { 3 } a + 9 z a$$

Factor

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

Solution Steps

Factor out z.

$$z\left(z^{3}-9z-z^{2}a+9a\right)$$

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

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

Factor out common term z^{2}-9 by using distributive property.

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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