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

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

Factor

$$-8\left(x-\frac{-\sqrt{257}-1}{16}\right)\left(x-\frac{\sqrt{257}-1}{16}\right)$$

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.

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

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.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-8\right)\times 8}}{2\left(-8\right)}$$

Multiply -4 times -8.

$$x=\frac{-\left(-1\right)±\sqrt{1+32\times 8}}{2\left(-8\right)}$$

Multiply 32 times 8.

$$x=\frac{-\left(-1\right)±\sqrt{1+256}}{2\left(-8\right)}$$

Add 1 to 256.

$$x=\frac{-\left(-1\right)±\sqrt{257}}{2\left(-8\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{257}}{2\left(-8\right)}$$

Multiply 2 times -8.

$$x=\frac{1±\sqrt{257}}{-16}$$

Now solve the equation x=\frac{1±\sqrt{257}}{-16} when ± is plus. Add 1 to \sqrt{257}.

$$x=\frac{\sqrt{257}+1}{-16}$$

Divide 1+\sqrt{257} by -16.

$$x=\frac{-\sqrt{257}-1}{16}$$

Now solve the equation x=\frac{1±\sqrt{257}}{-16} when ± is minus. Subtract \sqrt{257} from 1.

$$x=\frac{1-\sqrt{257}}{-16}$$

Divide 1-\sqrt{257} by -16.

$$x=\frac{\sqrt{257}-1}{16}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-1-\sqrt{257}}{16} for x_{1} and \frac{-1+\sqrt{257}}{16} for x_{2}.

$$-8x^{2}-x+8=-8\left(x-\frac{-\sqrt{257}-1}{16}\right)\left(x-\frac{\sqrt{257}-1}{16}\right)$$

Evaluate

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