How do you factor $\frac{1}{49} - x^{2}$ $1/49 - x^2$ ?

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How do you factor $\frac{1}{49} - x^{2}$ $1/49 - x^2$ ?

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Answer 1 · 2018-03-05T03:17:05

$(\frac{1}{49} - x^{2})$ $(1/49 - x^2)$ =
$(\frac{1}{7} + x) \cdot (\frac{1}{7} - x)$ $(1/7 + x) * (1/7 - x )$

Explanation:

$\frac{1}{49}$ $1/49$ is the same as ( $\frac{1^{2}}{7^{2}}$ $1^2/7^2$ ). $\frac{1}{49}$ $1/49$ and $x^{2}$ $x^2$ are both perfect squares so we use the difference of squares (factoring).
$(a^{2} - b^{2}) = (a + b) \cdot (a - b)$ $(a^2 - b^2) = (a + b) * (a - b)$
If you 'foil' the answer, you get $(a^{2} - a b + a b - b^{2})$ $(a^2 - ab +ab - b^2)$ . $(- a b)$ $(-ab)$ and $(+ a b)$ $(+ab)$ cancel, resulting in the original $(a^{2} - b^{2})$ $(a^2 - b^2)$ .
For the problem, the same process applies:
1. $(\frac{1}{49} - x^{2})$ $(1/49 - x^2)$
2. $(\frac{1}{7} + x) \cdot (\frac{1}{7} - x)$ $(1/7 + x) * (1/7 - x )$
To check, $[(\frac{1}{7} \cdot \frac{1}{7}) - (\frac{1}{7}) x + (\frac{1}{7}) x - (x \cdot x)] = (\frac{1}{49} - x^{2})$ $[(1/7 * 1/7) - (1/7)x + (1/7)x - (x*x)] = (1/49 - x^2)$

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How do you factor $\frac{1}{49} - x^{2}$ $1/49 - x^2$ ?

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