Solve (-x^2-4x-5)+(-x^2+8x+8)

$$( - x ^ { 2 } - 4 x - 5 ) + ( - x ^ { 2 } + 8 x + 8 )$$

Evaluate

$$3+4x-2x^{2}$$

Solution Steps

Combine -x^{2} and -x^{2} to get 2\left(-x^{2}\right).

$$2\left(-x^{2}\right)-4x-5+8x+8$$

Combine -4x and 8x to get 4x.

$$2\left(-x^{2}\right)+4x-5+8$$

Add -5 and 8 to get 3.

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

Multiply 2 and -1 to get -2.

$$-2x^{2}+4x+3$$

Factor

$$-2\left(x-\left(-\frac{\sqrt{10}}{2}+1\right)\right)\left(x-\left(\frac{\sqrt{10}}{2}+1\right)\right)$$

Steps Using the Quadratic Formula

Combine -x^{2} and -x^{2} to get 2\left(-x^{2}\right).

$$factor(2\left(-x^{2}\right)-4x-5+8x+8)$$

Combine -4x and 8x to get 4x.

$$factor(2\left(-x^{2}\right)+4x-5+8)$$

Add -5 and 8 to get 3.

$$factor(2\left(-x^{2}\right)+4x+3)$$

Multiply 2 and -1 to get -2.

$$factor(-2x^{2}+4x+3)$$

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.

$$-2x^{2}+4x+3=0$$

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.

$$x=\frac{-4±\sqrt{4^{2}-4\left(-2\right)\times 3}}{2\left(-2\right)}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-2\right)\times 3}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-4±\sqrt{16+8\times 3}}{2\left(-2\right)}$$

Multiply 8 times 3.

$$x=\frac{-4±\sqrt{16+24}}{2\left(-2\right)}$$

Add 16 to 24.

$$x=\frac{-4±\sqrt{40}}{2\left(-2\right)}$$

Take the square root of 40.

$$x=\frac{-4±2\sqrt{10}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-4±2\sqrt{10}}{-4}$$

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

$$x=\frac{2\sqrt{10}-4}{-4}$$

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

$$x=-\frac{\sqrt{10}}{2}+1$$

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

$$x=\frac{-2\sqrt{10}-4}{-4}$$

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

$$x=\frac{\sqrt{10}}{2}+1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 1-\frac{\sqrt{10}}{2} for x_{1} and 1+\frac{\sqrt{10}}{2} for x_{2}.

$$-2x^{2}+4x+3=-2\left(x-\left(-\frac{\sqrt{10}}{2}+1\right)\right)\left(x-\left(\frac{\sqrt{10}}{2}+1\right)\right)$$