Solve 6v^4+2v^3-20v^2

$$6 v ^ { 4 } + 2 v ^ { 3 } - 20 v ^ { 2 }$$

Factor

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

Solution Steps

Factor out 2.

$$2\left(3v^{4}+v^{3}-10v^{2}\right)$$

Consider 3v^{4}+v^{3}-10v^{2}. Factor out v^{2}.

$$v^{2}\left(3v^{2}+v-10\right)$$

Consider 3v^{2}+v-10. Factor the expression by grouping. First, the expression needs to be rewritten as 3v^{2}+av+bv-10. To find a and b, set up a system to be solved.

$$a+b=1 ab=3\left(-10\right)=-30$$

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 -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 1.

$$a=-5 b=6$$

Rewrite 3v^{2}+v-10 as \left(3v^{2}-5v\right)+\left(6v-10\right).

$$\left(3v^{2}-5v\right)+\left(6v-10\right)$$

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

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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