Solve f(x)=1/2x^2-5x+8

$$f ( x ) = \frac { 1 } { 2 } x ^ { 2 } - 5 x + 8$$

Factor

$$\frac{\left(x-8\right)\left(x-2\right)}{2}$$

Solution Steps

Factor out \frac{1}{2}.

$$\frac{x^{2}-10x+16}{2}$$

Consider x^{2}-10x+16. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+16. To find a and b, set up a system to be solved.

$$a+b=-10 ab=1\times 16=16$$

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 16.

$$-1,-16 -2,-8 -4,-4$$

Calculate the sum for each pair.

$$-1-16=-17 -2-8=-10 -4-4=-8$$

The solution is the pair that gives sum -10.

$$a=-8 b=-2$$

Rewrite x^{2}-10x+16 as \left(x^{2}-8x\right)+\left(-2x+16\right).

$$\left(x^{2}-8x\right)+\left(-2x+16\right)$$

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

$$x\left(x-8\right)-2\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(x-2\right)$$

Rewrite the complete factored expression.

$$\frac{\left(x-8\right)\left(x-2\right)}{2}$$

Evaluate

$$\frac{\left(x-8\right)\left(x-2\right)}{2}$$