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

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

Evaluate

$$v^{2}+10v+1$$

Solution Steps

Combine 3v and 7v to get 10v.

$$-v^{2}+10v-4+2v^{2}+5$$

Add -4 and 5 to get 1.

$$-v^{2}+10v+1+2v^{2}$$

Combine -v^{2} and 2v^{2} to get v^{2}.

$$v^{2}+10v+1$$

Factor

$$\left(v-\left(-2\sqrt{6}-5\right)\right)\left(v-\left(2\sqrt{6}-5\right)\right)$$

Steps Using the Quadratic Formula

Combine 3v and 7v to get 10v.

$$factor(-v^{2}+10v-4+2v^{2}+5)$$

Add -4 and 5 to get 1.

$$factor(-v^{2}+10v+1+2v^{2})$$

Combine -v^{2} and 2v^{2} to get v^{2}.

$$factor(v^{2}+10v+1)$$

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$v^{2}+10v+1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$v=\frac{-10±\sqrt{10^{2}-4}}{2}$$

Square 10.

$$v=\frac{-10±\sqrt{100-4}}{2}$$

Add 100 to -4.

$$v=\frac{-10±\sqrt{96}}{2}$$

Take the square root of 96.

$$v=\frac{-10±4\sqrt{6}}{2}$$

Now solve the equation v=\frac{-10±4\sqrt{6}}{2} when ± is plus. Add -10 to 4\sqrt{6}.

$$v=\frac{4\sqrt{6}-10}{2}$$

Divide -10+4\sqrt{6} by 2.

$$v=2\sqrt{6}-5$$

Now solve the equation v=\frac{-10±4\sqrt{6}}{2} when ± is minus. Subtract 4\sqrt{6} from -10.

$$v=\frac{-4\sqrt{6}-10}{2}$$

Divide -10-4\sqrt{6} by 2.

$$v=-2\sqrt{6}-5$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -5+2\sqrt{6} for x_{1} and -5-2\sqrt{6} for x_{2}.

$$v^{2}+10v+1=\left(v-\left(2\sqrt{6}-5\right)\right)\left(v-\left(-2\sqrt{6}-5\right)\right)$$