Solve 36v^3+66v^2-210v

$$36 v ^ { 3 } + 66 v ^ { 2 } - 210 v$$

Factor

$$6v\left(3v-5\right)\left(2v+7\right)$$

Solution Steps

Factor out 6.

$$6\left(6v^{3}+11v^{2}-35v\right)$$

Consider 6v^{3}+11v^{2}-35v. Factor out v.

$$v\left(6v^{2}+11v-35\right)$$

Consider 6v^{2}+11v-35. Factor the expression by grouping. First, the expression needs to be rewritten as 6v^{2}+av+bv-35. To find a and b, set up a system to be solved.

$$a+b=11 ab=6\left(-35\right)=-210$$

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 -210.

$$-1,210 -2,105 -3,70 -5,42 -6,35 -7,30 -10,21 -14,15$$

Calculate the sum for each pair.

$$-1+210=209 -2+105=103 -3+70=67 -5+42=37 -6+35=29 -7+30=23 -10+21=11 -14+15=1$$

The solution is the pair that gives sum 11.

$$a=-10 b=21$$

Rewrite 6v^{2}+11v-35 as \left(6v^{2}-10v\right)+\left(21v-35\right).

$$\left(6v^{2}-10v\right)+\left(21v-35\right)$$

Factor out 2v in the first and 7 in the second group.

$$2v\left(3v-5\right)+7\left(3v-5\right)$$

Factor out common term 3v-5 by using distributive property.

$$\left(3v-5\right)\left(2v+7\right)$$

Rewrite the complete factored expression.

$$6v\left(3v-5\right)\left(2v+7\right)$$

Evaluate

$$6v\left(3v-5\right)\left(2v+7\right)$$