Solve f(x)=x^2/6-x/3-1/2

$$f ( x ) = \frac { x ^ { 2 } } { 6 } - \frac { x } { 3 } - \frac { 1 } { 2 }$$

Evaluate

$$\frac{\left(x-3\right)\left(x+1\right)}{6}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 6 and 3 is 6. Multiply \frac{x}{3} times \frac{2}{2}.

$$\frac{x^{2}}{6}-\frac{2x}{6}-\frac{1}{2}$$

Since \frac{x^{2}}{6} and \frac{2x}{6} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x^{2}-2x}{6}-\frac{1}{2}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 6 and 2 is 6. Multiply \frac{1}{2} times \frac{3}{3}.

$$\frac{x^{2}-2x}{6}-\frac{3}{6}$$

Since \frac{x^{2}-2x}{6} and \frac{3}{6} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x^{2}-2x-3}{6}$$

Factor

$$\frac{\left(x-3\right)\left(x+1\right)}{6}$$

Solution Steps

Factor out \frac{1}{6}.

$$\frac{x^{2}-2x-3}{6}$$

Consider x^{2}-2x-3. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=-2 ab=1\left(-3\right)=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-3 b=1$$

Rewrite x^{2}-2x-3 as \left(x^{2}-3x\right)+\left(x-3\right).

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

Factor out x in x^{2}-3x.

$$x\left(x-3\right)+x-3$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(x+1\right)$$

Rewrite the complete factored expression.

$$\frac{\left(x-3\right)\left(x+1\right)}{6}$$