Solve f(x)=x^2/4-2x+3

$$f ( x ) = \frac { x ^ { 2 } } { 4 } - 2 x + 3$$

Evaluate

$$\frac{\left(x-6\right)\left(x-2\right)}{4}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply -2x+3 times \frac{4}{4}.

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

Since \frac{x^{2}}{4} and \frac{4\left(-2x+3\right)}{4} have the same denominator, add them by adding their numerators.

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

Do the multiplications in x^{2}+4\left(-2x+3\right).

$$\frac{x^{2}-8x+12}{4}$$

Factor

$$\frac{\left(x-6\right)\left(x-2\right)}{4}$$

Solution Steps

Factor out \frac{1}{4}.

$$\frac{x^{2}-8x+12}{4}$$

Consider x^{2}-8x+12. 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.

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

Rewrite the complete factored expression.

$$\frac{\left(x-6\right)\left(x-2\right)}{4}$$