hunterofdeath63

2021-12-18

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

nghodlokl

Expert

$20{x}^{5}-5=5\left(4{x}^{2}-1\right)=5\left(2x-1\right)\left(2x+1\right)$
Explanation:
So we have $20{x}^{2}-5=5\left(4{x}^{2}-1\right)$; dividing thru by 5
$5\left(4{x}^{2}-1\right)$
And $4{x}^{2}-1$ is the difference between 2 squares, $2x$ and 1
$\left(4{x}^{2}-1\right)=\left(2x-1\right)\left(2x+1\right)$
Putting it all together:
$20{x}^{2}-5=5\left(2x-1\right)\left(2x+1\right)$
So there are roots at $±\frac{1}{2}$

Travis Hicks

Expert

$20{x}^{2}-5$
Factor 5 out of $20{x}^{2}-5$
$5\left(4{x}^{2}-1\right)$
Rewrite $4{x}^{2}$ as ${\left(2x\right)}^{2}.$
$5\left({\left(2x\right)}^{2}-1\right)$
Rewrite $1$ as ${1}^{2}$
$5\left({\left(2x\right)}^{2}-{1}^{2}\right)$
$5\left(2x+1\right)\left(2x-1\right)$

RizerMix

Expert