Solve -16+16x+x^2

$$- 16 + 16 x + x ^ { 2 }$$

Factor

$$\left(x-\left(-4\sqrt{5}-8\right)\right)\left(x-\left(4\sqrt{5}-8\right)\right)$$

Steps Using the Quadratic Formula

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.

$$x^{2}+16x-16=0$$

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.

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

Square 16.

$$x=\frac{-16±\sqrt{256-4\left(-16\right)}}{2}$$

Multiply -4 times -16.

$$x=\frac{-16±\sqrt{256+64}}{2}$$

Add 256 to 64.

$$x=\frac{-16±\sqrt{320}}{2}$$

Take the square root of 320.

$$x=\frac{-16±8\sqrt{5}}{2}$$

Now solve the equation x=\frac{-16±8\sqrt{5}}{2} when ± is plus. Add -16 to 8\sqrt{5}.

$$x=\frac{8\sqrt{5}-16}{2}$$

Divide -16+8\sqrt{5} by 2.

$$x=4\sqrt{5}-8$$

Now solve the equation x=\frac{-16±8\sqrt{5}}{2} when ± is minus. Subtract 8\sqrt{5} from -16.

$$x=\frac{-8\sqrt{5}-16}{2}$$

Divide -16-8\sqrt{5} by 2.

$$x=-4\sqrt{5}-8$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -8+4\sqrt{5} for x_{1} and -8-4\sqrt{5} for x_{2}.

$$x^{2}+16x-16=\left(x-\left(4\sqrt{5}-8\right)\right)\left(x-\left(-4\sqrt{5}-8\right)\right)$$

Evaluate

$$x^{2}+16x-16$$