Solve 36x^3+30x^2-24x

$$36 { x }^{ 3 } +30 { x }^{ 2 } -24x$$

Factor

$$6x\left(2x-1\right)\left(3x+4\right)$$

Solution Steps

Factor out 6.

$$6\left(6x^{3}+5x^{2}-4x\right)$$

Consider 6x^{3}+5x^{2}-4x. Factor out x.

$$x\left(6x^{2}+5x-4\right)$$

Consider 6x^{2}+5x-4. Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=5 ab=6\left(-4\right)=-24$$

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 5.

$$a=-3 b=8$$

Rewrite 6x^{2}+5x-4 as \left(6x^{2}-3x\right)+\left(8x-4\right).

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

Factor out 3x in the first and 4 in the second group.

$$3x\left(2x-1\right)+4\left(2x-1\right)$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(3x+4\right)$$

Rewrite the complete factored expression.

$$6x\left(2x-1\right)\left(3x+4\right)$$

Evaluate

$$6x\left(2x-1\right)\left(3x+4\right)$$