Solve z^3-6z^2-72z

$$z ^ { 3 } - 6 z ^ { 2 } - 72 z$$

Factor

$$z\left(z-12\right)\left(z+6\right)$$

Solution Steps

Factor out z.

$$z\left(z^{2}-6z-72\right)$$

Consider z^{2}-6z-72. Factor the expression by grouping. First, the expression needs to be rewritten as z^{2}+az+bz-72. To find a and b, set up a system to be solved.

$$a+b=-6 ab=1\left(-72\right)=-72$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -72.

$$1,-72 2,-36 3,-24 4,-18 6,-12 8,-9$$

Calculate the sum for each pair.

$$1-72=-71 2-36=-34 3-24=-21 4-18=-14 6-12=-6 8-9=-1$$

The solution is the pair that gives sum -6.

$$a=-12 b=6$$

Rewrite z^{2}-6z-72 as \left(z^{2}-12z\right)+\left(6z-72\right).

$$\left(z^{2}-12z\right)+\left(6z-72\right)$$

Factor out z in the first and 6 in the second group.

$$z\left(z-12\right)+6\left(z-12\right)$$

Factor out common term z-12 by using distributive property.

$$\left(z-12\right)\left(z+6\right)$$

Rewrite the complete factored expression.

$$z\left(z-12\right)\left(z+6\right)$$

Evaluate

$$z\left(z-12\right)\left(z+6\right)$$