How do you factor $20x^2 - 5$ ?

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Answer 1 · 2016-08-29T09:14:43

$20x^2-5=5(4x^(2)-1)$ $=$ $5(2x-1)(2x+1)$

So we have $20x^2-5=5(4x^(2)-1)$ ; dividing thru by $5$ .

$5(4x^(2)-1)$

And $4x^2-1$ is the difference between 2 squares, $2x$ and $1$ .

$(4x^(2)-1)$ $=$ $(2x-1)(2x+1)$

Putting it all together:

$20x^2-5=5(2x-1)(2x+1)$

So there are roots at $+-1/2$ .

How do you factor 20x^2 - 520x^2 - 5?