Solve x^2/2-1/2+2x-2

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

Evaluate

$$\frac{\left(x-1\right)\left(x+5\right)}{2}$$

Short Solution Steps

Subtract 2 from -\frac{1}{2} to get -\frac{5}{2}.

$$\frac{x^{2}}{2}-\frac{5}{2}+2x$$

Since \frac{x^{2}}{2} and \frac{5}{2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x^{2}-5}{2}+2x$$

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

$$\frac{x^{2}-5}{2}+\frac{2\times 2x}{2}$$

Since \frac{x^{2}-5}{2} and \frac{2\times 2x}{2} have the same denominator, add them by adding their numerators.

$$\frac{x^{2}-5+2\times 2x}{2}$$

Do the multiplications in x^{2}-5+2\times 2x.

$$\frac{x^{2}-5+4x}{2}$$

Factor

$$\frac{\left(x-1\right)\left(x+5\right)}{2}$$

Solution Steps

Factor out \frac{1}{2}.

$$\frac{x^{2}-5+4x}{2}$$

Consider x^{2}-1+4x-4. Multiply and combine like terms.

$$x^{2}+4x-5$$

Consider x^{2}+4x-5. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\left(-5\right)=-5$$

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

$$a=-1 b=5$$

Rewrite x^{2}+4x-5 as \left(x^{2}-x\right)+\left(5x-5\right).

$$\left(x^{2}-x\right)+\left(5x-5\right)$$

Factor out x in the first and 5 in the second group.

$$x\left(x-1\right)+5\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(x+5\right)$$

Rewrite the complete factored expression.

$$\frac{\left(x-1\right)\left(x+5\right)}{2}$$