Solve x^2-8x+12

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

Factor

$$\left(x-6\right)\left(x-2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=-8 ab=1\times 12=12$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -8.

$$a=-6 b=-2$$

Rewrite x^{2}-8x+12 as \left(x^{2}-6x\right)+\left(-2x+12\right).

$$\left(x^{2}-6x\right)+\left(-2x+12\right)$$

Factor out x in the first and -2 in the second group.

$$x\left(x-6\right)-2\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(x-2\right)$$

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.

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

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.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 12}}{2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 12}}{2}$$

Multiply -4 times 12.

$$x=\frac{-\left(-8\right)±\sqrt{64-48}}{2}$$

Add 64 to -48.

$$x=\frac{-\left(-8\right)±\sqrt{16}}{2}$$

Take the square root of 16.

$$x=\frac{-\left(-8\right)±4}{2}$$

The opposite of -8 is 8.

$$x=\frac{8±4}{2}$$

Now solve the equation x=\frac{8±4}{2} when ± is plus. Add 8 to 4.

$$x=\frac{12}{2}$$

Divide 12 by 2.

$$x=6$$

Now solve the equation x=\frac{8±4}{2} when ± is minus. Subtract 4 from 8.

$$x=\frac{4}{2}$$

Divide 4 by 2.

$$x=2$$

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

$$x^{2}-8x+12=\left(x-6\right)\left(x-2\right)$$

Evaluate

$$\left(x-6\right)\left(x-2\right)$$