Solve (-3v^2+4v-5)+(7v^2+2v+1)

$$( - 3 v ^ { 2 } + 4 v - 5 ) + ( 7 v ^ { 2 } + 2 v + 1 )$$

Evaluate

$$2\left(2v-1\right)\left(v+2\right)$$

Solution Steps

Combine -3v^{2} and 7v^{2} to get 4v^{2}.

$$4v^{2}+4v-5+2v+1$$

Combine 4v and 2v to get 6v.

$$4v^{2}+6v-5+1$$

Add -5 and 1 to get -4.

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

Factor

$$2\left(2v-1\right)\left(v+2\right)$$

Solution Steps

Multiply and combine like terms.

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

Factor out 2.

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

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

$$a+b=3 ab=2\left(-2\right)=-4$$

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 -4.

$$-1,4 -2,2$$

Calculate the sum for each pair.

$$-1+4=3 -2+2=0$$

The solution is the pair that gives sum 3.

$$a=-1 b=4$$

Rewrite 2v^{2}+3v-2 as \left(2v^{2}-v\right)+\left(4v-2\right).

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

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

$$v\left(2v-1\right)+2\left(2v-1\right)$$

Factor out common term 2v-1 by using distributive property.

$$\left(2v-1\right)\left(v+2\right)$$

Rewrite the complete factored expression.

$$2\left(2v-1\right)\left(v+2\right)$$