Solve 1/8x^2-1/96x-1/16

$$\frac{ 1 }{ 8 } { x }^{ 2 } - \frac{ 1 }{ 96 } x- \frac{ 1 }{ 16 }$$

Factor

$$\frac{\left(4x-3\right)\left(3x+2\right)}{96}$$

Solution Steps

Factor out \frac{1}{96}.

$$\frac{12x^{2}-x-6}{96}$$

Consider 12x^{2}-x-6. Factor the expression by grouping. First, the expression needs to be rewritten as 12x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=-1 ab=12\left(-6\right)=-72$$

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. List all such integer pairs that give product -72.

$$1,-72 2,-36 3,-24 4,-18 6,-12 8,-9$$

Calculate the sum for each pair.

$$1-72=-71 2-36=-34 3-24=-21 4-18=-14 6-12=-6 8-9=-1$$

The solution is the pair that gives sum -1.

$$a=-9 b=8$$

Rewrite 12x^{2}-x-6 as \left(12x^{2}-9x\right)+\left(8x-6\right).

$$\left(12x^{2}-9x\right)+\left(8x-6\right)$$

Factor out 3x in the first and 2 in the second group.

$$3x\left(4x-3\right)+2\left(4x-3\right)$$

Factor out common term 4x-3 by using distributive property.

$$\left(4x-3\right)\left(3x+2\right)$$

Rewrite the complete factored expression.

$$\frac{\left(4x-3\right)\left(3x+2\right)}{96}$$

Evaluate

$$\frac{x^{2}}{8}-\frac{x}{96}-\frac{1}{16}$$