Multiply by using the special case products. $$\left( x - \frac { 1 } { 8 } \right) \left( x + \frac { 1 } { 8 } \right)$$
A) $x ^ { 2 } + \frac { 2 } { 25 } x - \frac { 1 } { 64 }$
B) $x ^ { 2 } - \frac { 2 } { 25 } x - \frac { 1 } { 64 }$
C) $x ^ { 2 } + \frac { 1 } { 64 }$
D) $x ^ { 2 } - \frac { 1 } { 64 }$

Question

Quizplus · Accepted Answer

This is an application of the difference of squares formula, $a^2 - b^2 = (a - b)(a + b)$, where $a = x$ and $b = \frac{1}{8}$, resulting in $x^2 - \left(\frac{1}{8}\right)^2 = x^2 - \frac{1}{64}$.

Multiply by Using the Special Case Products (x−18)(x+18)\left( x - \frac { 1 } { 8 } \right) \left( x + \frac { 1 } { 8 } \right)(x−81​)(x+81​)

Multiply by Using the Special Case Products $\left( x - \frac { 1 } { 8 } \right) \left( x + \frac { 1 } { 8 } \right)$