Solve 4x^2-12x+4

$$4 { x }^{ 2 } -12x+4$$

Factor

$$4\left(x-\frac{3-\sqrt{5}}{2}\right)\left(x-\frac{\sqrt{5}+3}{2}\right)$$

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.

$$4x^{2}-12x+4=0$$

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.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 4\times 4}}{2\times 4}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 4\times 4}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-12\right)±\sqrt{144-16\times 4}}{2\times 4}$$

Multiply -16 times 4.

$$x=\frac{-\left(-12\right)±\sqrt{144-64}}{2\times 4}$$

Add 144 to -64.

$$x=\frac{-\left(-12\right)±\sqrt{80}}{2\times 4}$$

Take the square root of 80.

$$x=\frac{-\left(-12\right)±4\sqrt{5}}{2\times 4}$$

The opposite of -12 is 12.

$$x=\frac{12±4\sqrt{5}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{12±4\sqrt{5}}{8}$$

Now solve the equation x=\frac{12±4\sqrt{5}}{8} when ± is plus. Add 12 to 4\sqrt{5}.

$$x=\frac{4\sqrt{5}+12}{8}$$

Divide 12+4\sqrt{5} by 8.

$$x=\frac{\sqrt{5}+3}{2}$$

Now solve the equation x=\frac{12±4\sqrt{5}}{8} when ± is minus. Subtract 4\sqrt{5} from 12.

$$x=\frac{12-4\sqrt{5}}{8}$$

Divide 12-4\sqrt{5} by 8.

$$x=\frac{3-\sqrt{5}}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3+\sqrt{5}}{2} for x_{1} and \frac{3-\sqrt{5}}{2} for x_{2}.

$$4x^{2}-12x+4=4\left(x-\frac{\sqrt{5}+3}{2}\right)\left(x-\frac{3-\sqrt{5}}{2}\right)$$

Evaluate

$$4\left(x^{2}-3x+1\right)$$