Solve -1/2a^2+2a-2

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

Factor

$$-\frac{\left(a-2\right)^{2}}{2}$$

Solution Steps

Factor out \frac{1}{2}.

$$\frac{-a^{2}+4a-4}{2}$$

Consider -a^{2}+4a-4. Factor the expression by grouping. First, the expression needs to be rewritten as -a^{2}+pa+qa-4. To find p and q, set up a system to be solved.

$$p+q=4 pq=-\left(-4\right)=4$$

Since pq is positive, p and q have the same sign. Since p+q is positive, p and q are both positive. List all such integer pairs that give product 4.

$$1,4 2,2$$

Calculate the sum for each pair.

$$1+4=5 2+2=4$$

The solution is the pair that gives sum 4.

$$p=2 q=2$$

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

$$\left(-a^{2}+2a\right)+\left(2a-4\right)$$

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

$$-a\left(a-2\right)+2\left(a-2\right)$$

Factor out common term a-2 by using distributive property.

$$\left(a-2\right)\left(-a+2\right)$$

Rewrite the complete factored expression.

$$\frac{\left(a-2\right)\left(-a+2\right)}{2}$$

Evaluate

$$-\frac{\left(a-2\right)^{2}}{2}$$