Solve 3x^2+8x+2

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

Factor

$$3\left(x-\frac{-\sqrt{10}-4}{3}\right)\left(x-\frac{\sqrt{10}-4}{3}\right)$$

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.

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

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.

$$x=\frac{-8±\sqrt{8^{2}-4\times 3\times 2}}{2\times 3}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 3\times 2}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-8±\sqrt{64-12\times 2}}{2\times 3}$$

Multiply -12 times 2.

$$x=\frac{-8±\sqrt{64-24}}{2\times 3}$$

Add 64 to -24.

$$x=\frac{-8±\sqrt{40}}{2\times 3}$$

Take the square root of 40.

$$x=\frac{-8±2\sqrt{10}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-8±2\sqrt{10}}{6}$$

Now solve the equation x=\frac{-8±2\sqrt{10}}{6} when ± is plus. Add -8 to 2\sqrt{10}.

$$x=\frac{2\sqrt{10}-8}{6}$$

Divide -8+2\sqrt{10} by 6.

$$x=\frac{\sqrt{10}-4}{3}$$

Now solve the equation x=\frac{-8±2\sqrt{10}}{6} when ± is minus. Subtract 2\sqrt{10} from -8.

$$x=\frac{-2\sqrt{10}-8}{6}$$

Divide -8-2\sqrt{10} by 6.

$$x=\frac{-\sqrt{10}-4}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-4+\sqrt{10}}{3} for x_{1} and \frac{-4-\sqrt{10}}{3} for x_{2}.

$$3x^{2}+8x+2=3\left(x-\frac{\sqrt{10}-4}{3}\right)\left(x-\frac{-\sqrt{10}-4}{3}\right)$$

Evaluate

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