Solve 9v^3+15v^2-36v

$$9 v ^ { 3 } + 15 v ^ { 2 } - 36 v$$

Factor

$$3v\left(3v-4\right)\left(v+3\right)$$

Solution Steps

Factor out 3.

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

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

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

Consider 3v^{2}+5v-12. Factor the expression by grouping. First, the expression needs to be rewritten as 3v^{2}+av+bv-12. To find a and b, set up a system to be solved.

$$a+b=5 ab=3\left(-12\right)=-36$$

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 -36.

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

Calculate the sum for each pair.

$$-1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0$$

The solution is the pair that gives sum 5.

$$a=-4 b=9$$

Rewrite 3v^{2}+5v-12 as \left(3v^{2}-4v\right)+\left(9v-12\right).

$$\left(3v^{2}-4v\right)+\left(9v-12\right)$$

Factor out v in the first and 3 in the second group.

$$v\left(3v-4\right)+3\left(3v-4\right)$$

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

$$\left(3v-4\right)\left(v+3\right)$$

Rewrite the complete factored expression.

$$3v\left(3v-4\right)\left(v+3\right)$$

Evaluate

$$3v\left(3v-4\right)\left(v+3\right)$$