Solve 3x^2+12x+8

$$3 { x }^{ 2 } +12x+8$$

Factor

$$3\left(x-\left(-\frac{2\sqrt{3}}{3}-2\right)\right)\left(x-\left(\frac{2\sqrt{3}}{3}-2\right)\right)$$

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.

$$3x^{2}+12x+8=0$$

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.

$$x=\frac{-12±\sqrt{12^{2}-4\times 3\times 8}}{2\times 3}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 3\times 8}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-12±\sqrt{144-12\times 8}}{2\times 3}$$

Multiply -12 times 8.

$$x=\frac{-12±\sqrt{144-96}}{2\times 3}$$

Add 144 to -96.

$$x=\frac{-12±\sqrt{48}}{2\times 3}$$

Take the square root of 48.

$$x=\frac{-12±4\sqrt{3}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-12±4\sqrt{3}}{6}$$

Now solve the equation x=\frac{-12±4\sqrt{3}}{6} when ± is plus. Add -12 to 4\sqrt{3}.

$$x=\frac{4\sqrt{3}-12}{6}$$

Divide -12+4\sqrt{3} by 6.

$$x=\frac{2\sqrt{3}}{3}-2$$

Now solve the equation x=\frac{-12±4\sqrt{3}}{6} when ± is minus. Subtract 4\sqrt{3} from -12.

$$x=\frac{-4\sqrt{3}-12}{6}$$

Divide -12-4\sqrt{3} by 6.

$$x=-\frac{2\sqrt{3}}{3}-2$$

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

$$3x^{2}+12x+8=3\left(x-\left(\frac{2\sqrt{3}}{3}-2\right)\right)\left(x-\left(-\frac{2\sqrt{3}}{3}-2\right)\right)$$

Evaluate

$$3x^{2}+12x+8$$