Solve 10v^7-32v^6+6v^5

$$10 v ^ { 7 } - 32 v ^ { 6 } + 6 v ^ { 5 }$$

Factor

$$2\left(v-3\right)\left(5v-1\right)v^{5}$$

Solution Steps

Factor out 2.

$$2\left(5v^{7}-16v^{6}+3v^{5}\right)$$

Consider 5v^{7}-16v^{6}+3v^{5}. Factor out v^{5}.

$$v^{5}\left(5v^{2}-16v+3\right)$$

Consider 5v^{2}-16v+3. Factor the expression by grouping. First, the expression needs to be rewritten as 5v^{2}+av+bv+3. To find a and b, set up a system to be solved.

$$a+b=-16 ab=5\times 3=15$$

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 15.

$$-1,-15 -3,-5$$

Calculate the sum for each pair.

$$-1-15=-16 -3-5=-8$$

The solution is the pair that gives sum -16.

$$a=-15 b=-1$$

Rewrite 5v^{2}-16v+3 as \left(5v^{2}-15v\right)+\left(-v+3\right).

$$\left(5v^{2}-15v\right)+\left(-v+3\right)$$

Factor out 5v in the first and -1 in the second group.

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

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

$$\left(v-3\right)\left(5v-1\right)$$

Rewrite the complete factored expression.

$$2v^{5}\left(v-3\right)\left(5v-1\right)$$

Evaluate

$$2\left(v-3\right)\left(5v-1\right)v^{5}$$