Solve (x^2-8x-4)

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

Factor

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

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.

$$x^{2}-8x-4=0$$

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.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-4\right)}}{2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\left(-4\right)}}{2}$$

Multiply -4 times -4.

$$x=\frac{-\left(-8\right)±\sqrt{64+16}}{2}$$

Add 64 to 16.

$$x=\frac{-\left(-8\right)±\sqrt{80}}{2}$$

Take the square root of 80.

$$x=\frac{-\left(-8\right)±4\sqrt{5}}{2}$$

The opposite of -8 is 8.

$$x=\frac{8±4\sqrt{5}}{2}$$

Now solve the equation x=\frac{8±4\sqrt{5}}{2} when ± is plus. Add 8 to 4\sqrt{5}.

$$x=\frac{4\sqrt{5}+8}{2}$$

Divide 8+4\sqrt{5} by 2.

$$x=2\sqrt{5}+4$$

Now solve the equation x=\frac{8±4\sqrt{5}}{2} when ± is minus. Subtract 4\sqrt{5} from 8.

$$x=\frac{8-4\sqrt{5}}{2}$$

Divide 8-4\sqrt{5} by 2.

$$x=4-2\sqrt{5}$$

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

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

Evaluate

$$x^{2}-8x-4$$