Solve 9d^3+6d^2-24d

$$9 d ^ { 3 } + 6 d ^ { 2 } - 24 d$$

Factor

$$3d\left(3d-4\right)\left(d+2\right)$$

Solution Steps

Factor out 3.

$$3\left(3d^{3}+2d^{2}-8d\right)$$

Consider 3d^{3}+2d^{2}-8d. Factor out d.

$$d\left(3d^{2}+2d-8\right)$$

Consider 3d^{2}+2d-8. Factor the expression by grouping. First, the expression needs to be rewritten as 3d^{2}+ad+bd-8. To find a and b, set up a system to be solved.

$$a+b=2 ab=3\left(-8\right)=-24$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -24.

$$-1,24 -2,12 -3,8 -4,6$$

Calculate the sum for each pair.

$$-1+24=23 -2+12=10 -3+8=5 -4+6=2$$

The solution is the pair that gives sum 2.

$$a=-4 b=6$$

Rewrite 3d^{2}+2d-8 as \left(3d^{2}-4d\right)+\left(6d-8\right).

$$\left(3d^{2}-4d\right)+\left(6d-8\right)$$

Factor out d in the first and 2 in the second group.

$$d\left(3d-4\right)+2\left(3d-4\right)$$

Factor out common term 3d-4 by using distributive property.

$$\left(3d-4\right)\left(d+2\right)$$

Rewrite the complete factored expression.

$$3d\left(3d-4\right)\left(d+2\right)$$

Evaluate

$$3d\left(3d-4\right)\left(d+2\right)$$