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

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Answer 1 · 2016-04-09T18:01:19

$(\frac{x}{2} + 2) (\frac{x}{2} - 2)$ $(x/2+2)(x/2-2)$

We will use the following property to find the factors of the given expression:

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

Solution:

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

Here, $a^{2} = \frac{1}{4} x^{2}$ $a^2=1/4x^2$ and $b^{2} = 4$ $b^2=4$

$\Rightarrow a = \frac{1}{2} x$ $=>a=1/2x$ and $b = 2$ $b = 2$

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

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