Solve 112v^3+34v^2-6v

$$112 v ^ { 3 } + 34 v ^ { 2 } - 6 v$$

Factor

$$2v\left(8v-1\right)\left(7v+3\right)$$

Solution Steps

Factor out 2.

$$2\left(56v^{3}+17v^{2}-3v\right)$$

Consider 56v^{3}+17v^{2}-3v. Factor out v.

$$v\left(56v^{2}+17v-3\right)$$

Consider 56v^{2}+17v-3. Factor the expression by grouping. First, the expression needs to be rewritten as 56v^{2}+av+bv-3. To find a and b, set up a system to be solved.

$$a+b=17 ab=56\left(-3\right)=-168$$

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 -168.

$$-1,168 -2,84 -3,56 -4,42 -6,28 -7,24 -8,21 -12,14$$

Calculate the sum for each pair.

$$-1+168=167 -2+84=82 -3+56=53 -4+42=38 -6+28=22 -7+24=17 -8+21=13 -12+14=2$$

The solution is the pair that gives sum 17.

$$a=-7 b=24$$

Rewrite 56v^{2}+17v-3 as \left(56v^{2}-7v\right)+\left(24v-3\right).

$$\left(56v^{2}-7v\right)+\left(24v-3\right)$$

Factor out 7v in the first and 3 in the second group.

$$7v\left(8v-1\right)+3\left(8v-1\right)$$

Factor out common term 8v-1 by using distributive property.

$$\left(8v-1\right)\left(7v+3\right)$$

Rewrite the complete factored expression.

$$2v\left(8v-1\right)\left(7v+3\right)$$

Evaluate

$$2v\left(8v-1\right)\left(7v+3\right)$$