How do you factor the expression $\frac{1}{9} x^{2} - \frac{1}{4}$ $1/9x^2 - 1/4$ ?

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How do you factor the expression $\frac{1}{9} x^{2} - \frac{1}{4}$ $1/9x^2 - 1/4$ ?

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Answer 1 · 2015-12-27T12:21:51

Use the difference of squares identity to find:

$\frac{1}{9} x^{2} - \frac{1}{4} = (\frac{x}{3} - \frac{1}{2}) (\frac{x}{3} + \frac{1}{2})$ $1/9x^2-1/4 = (x/3-1/2)(x/3+1/2)$

Explanation:

Both $\frac{1}{9} x^{2} = {(\frac{x}{3})}^{2}$ $1/9 x^2 = (x/3)^2$ and $\frac{1}{4} = {(\frac{1}{2})}^{2}$ $1/4 = (1/2)^2$ are perfect squares.

So you can conveniently use the difference of squares identity:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2 = (a-b)(a+b)$

with $a = \frac{x}{3}$ $a = x/3$ and $b = \frac{1}{2}$ $b=1/2$ to find:

$\frac{1}{9} x^{2} - \frac{1}{4} = {(\frac{x}{3})}^{2} - {(\frac{1}{2})}^{2} = (\frac{x}{3} - \frac{1}{2}) (\frac{x}{3} + \frac{1}{2})$ $1/9x^2-1/4 = (x/3)^2-(1/2)^2 = (x/3-1/2)(x/3+1/2)$

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How do you factor the expression $\frac{1}{9} x^{2} - \frac{1}{4}$ $1/9x^2 - 1/4$ ?

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