Solve 3x^2-x-8

$$3 { x }^{ 2 } -x-8$$

Factor

$$3\left(x-\frac{1-\sqrt{97}}{6}\right)\left(x-\frac{\sqrt{97}+1}{6}\right)$$

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.

$$3x^{2}-x-8=0$$

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.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 3\left(-8\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-1\right)±\sqrt{1-12\left(-8\right)}}{2\times 3}$$

Multiply -12 times -8.

$$x=\frac{-\left(-1\right)±\sqrt{1+96}}{2\times 3}$$

Add 1 to 96.

$$x=\frac{-\left(-1\right)±\sqrt{97}}{2\times 3}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{97}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{1±\sqrt{97}}{6}$$

Now solve the equation x=\frac{1±\sqrt{97}}{6} when ± is plus. Add 1 to \sqrt{97}.

$$x=\frac{\sqrt{97}+1}{6}$$

Now solve the equation x=\frac{1±\sqrt{97}}{6} when ± is minus. Subtract \sqrt{97} from 1.

$$x=\frac{1-\sqrt{97}}{6}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1+\sqrt{97}}{6} for x_{1} and \frac{1-\sqrt{97}}{6} for x_{2}.

$$3x^{2}-x-8=3\left(x-\frac{\sqrt{97}+1}{6}\right)\left(x-\frac{1-\sqrt{97}}{6}\right)$$

Evaluate

$$3x^{2}-x-8$$