Solve -1/4x^2-2x-3

$$- \frac { 1 } { 4 } x ^ { 2 } - 2 x - 3$$

Factor

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

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.

$$a+b=-8 ab=-\left(-12\right)=12$$

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.

$$a=-2 b=-6$$

Rewrite -x^{2}-8x-12 as \left(-x^{2}-2x\right)+\left(-6x-12\right).

$$\left(-x^{2}-2x\right)+\left(-6x-12\right)$$

Factor out x in the first and 6 in the second group.

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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