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

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Answer 1 · 2016-08-29T09:14:43

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

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

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

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

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

Putting it all together:

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

So there are roots at $\pm \frac{1}{2}$ $+-1/2$ .

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