Solve 2v^4-22v^3+48v^2

$$2 v ^ { 4 } - 22 v ^ { 3 } + 48 v ^ { 2 }$$

Factor

$$2\left(v-8\right)\left(v-3\right)v^{2}$$

Solution Steps

Factor out 2.

$$2\left(v^{4}-11v^{3}+24v^{2}\right)$$

Consider v^{4}-11v^{3}+24v^{2}. Factor out v^{2}.

$$v^{2}\left(v^{2}-11v+24\right)$$

Consider v^{2}-11v+24. Factor the expression by grouping. First, the expression needs to be rewritten as v^{2}+av+bv+24. To find a and b, set up a system to be solved.

$$a+b=-11 ab=1\times 24=24$$

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 24.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Calculate the sum for each pair.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

The solution is the pair that gives sum -11.

$$a=-8 b=-3$$

Rewrite v^{2}-11v+24 as \left(v^{2}-8v\right)+\left(-3v+24\right).

$$\left(v^{2}-8v\right)+\left(-3v+24\right)$$

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

$$v\left(v-8\right)-3\left(v-8\right)$$

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

$$\left(v-8\right)\left(v-3\right)$$

Rewrite the complete factored expression.

$$2v^{2}\left(v-8\right)\left(v-3\right)$$

Evaluate

$$2\left(v-8\right)\left(v-3\right)v^{2}$$