Solve 6z^3-18z^2-24z

$$6 z ^ { 3 } - 18 z ^ { 2 } - 24 z$$

Factor

$$6z\left(z-4\right)\left(z+1\right)$$

Solution Steps

Factor out 6.

$$6\left(z^{3}-3z^{2}-4z\right)$$

Consider z^{3}-3z^{2}-4z. Factor out z.

$$z\left(z^{2}-3z-4\right)$$

Consider z^{2}-3z-4. Factor the expression by grouping. First, the expression needs to be rewritten as z^{2}+az+bz-4. To find a and b, set up a system to be solved.

$$a+b=-3 ab=1\left(-4\right)=-4$$

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 -4.

$$1,-4 2,-2$$

Calculate the sum for each pair.

$$1-4=-3 2-2=0$$

The solution is the pair that gives sum -3.

$$a=-4 b=1$$

Rewrite z^{2}-3z-4 as \left(z^{2}-4z\right)+\left(z-4\right).

$$\left(z^{2}-4z\right)+\left(z-4\right)$$

Factor out z in z^{2}-4z.

$$z\left(z-4\right)+z-4$$

Factor out common term z-4 by using distributive property.

$$\left(z-4\right)\left(z+1\right)$$

Rewrite the complete factored expression.

$$6z\left(z-4\right)\left(z+1\right)$$

Evaluate

$$6z\left(z-4\right)\left(z+1\right)$$