Solve 6x^3-15x^2-36x

$$6 x ^ { 3 } - 15 x ^ { 2 } - 36 x$$

Factor

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

Solution Steps

Factor out 3.

$$3\left(2x^{3}-5x^{2}-12x\right)$$

Consider 2x^{3}-5x^{2}-12x. Factor out x.

$$x\left(2x^{2}-5x-12\right)$$

Consider 2x^{2}-5x-12. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=-5 ab=2\left(-12\right)=-24$$

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 -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.

$$a=-8 b=3$$

Rewrite 2x^{2}-5x-12 as \left(2x^{2}-8x\right)+\left(3x-12\right).

$$\left(2x^{2}-8x\right)+\left(3x-12\right)$$

Factor out 2x in the first and 3 in the second group.

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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