Solve 3x^2-9x+3

$$3 { x }^{ 2 } -9x+3$$

Factor

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

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.

$$3x^{2}-9x+3=0$$

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.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 3\times 3}}{2\times 3}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 3\times 3}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-9\right)±\sqrt{81-12\times 3}}{2\times 3}$$

Multiply -12 times 3.

$$x=\frac{-\left(-9\right)±\sqrt{81-36}}{2\times 3}$$

Add 81 to -36.

$$x=\frac{-\left(-9\right)±\sqrt{45}}{2\times 3}$$

Take the square root of 45.

$$x=\frac{-\left(-9\right)±3\sqrt{5}}{2\times 3}$$

The opposite of -9 is 9.

$$x=\frac{9±3\sqrt{5}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{9±3\sqrt{5}}{6}$$

Now solve the equation x=\frac{9±3\sqrt{5}}{6} when ± is plus. Add 9 to 3\sqrt{5}.

$$x=\frac{3\sqrt{5}+9}{6}$$

Divide 9+3\sqrt{5} by 6.

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

Now solve the equation x=\frac{9±3\sqrt{5}}{6} when ± is minus. Subtract 3\sqrt{5} from 9.

$$x=\frac{9-3\sqrt{5}}{6}$$

Divide 9-3\sqrt{5} by 6.

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

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}.

$$3x^{2}-9x+3=3\left(x-\frac{\sqrt{5}+3}{2}\right)\left(x-\frac{3-\sqrt{5}}{2}\right)$$

Evaluate

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