Solve 360v^3-702v^2+324v

$$360 v ^ { 3 } - 702 v ^ { 2 } + 324 v$$

Factor

$$18v\left(4v-3\right)\left(5v-6\right)$$

Solution Steps

Factor out 18.

$$18\left(20v^{3}-39v^{2}+18v\right)$$

Consider 20v^{3}-39v^{2}+18v. Factor out v.

$$v\left(20v^{2}-39v+18\right)$$

Consider 20v^{2}-39v+18. Factor the expression by grouping. First, the expression needs to be rewritten as 20v^{2}+av+bv+18. To find a and b, set up a system to be solved.

$$a+b=-39 ab=20\times 18=360$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 360.

$$-1,-360 -2,-180 -3,-120 -4,-90 -5,-72 -6,-60 -8,-45 -9,-40 -10,-36 -12,-30 -15,-24 -18,-20$$

Calculate the sum for each pair.

$$-1-360=-361 -2-180=-182 -3-120=-123 -4-90=-94 -5-72=-77 -6-60=-66 -8-45=-53 -9-40=-49 -10-36=-46 -12-30=-42 -15-24=-39 -18-20=-38$$

The solution is the pair that gives sum -39.

$$a=-24 b=-15$$

Rewrite 20v^{2}-39v+18 as \left(20v^{2}-24v\right)+\left(-15v+18\right).

$$\left(20v^{2}-24v\right)+\left(-15v+18\right)$$

Factor out 4v in the first and -3 in the second group.

$$4v\left(5v-6\right)-3\left(5v-6\right)$$

Factor out common term 5v-6 by using distributive property.

$$\left(5v-6\right)\left(4v-3\right)$$

Rewrite the complete factored expression.

$$18v\left(5v-6\right)\left(4v-3\right)$$

Evaluate

$$18v\left(4v-3\right)\left(5v-6\right)$$