Question
Factorise:
$16 x^4-1$
$16 x^4-1$
✓
Answer
$16 x^4-1$
$=\left(4 x^2\right)^2-(1)^2$
$=\left(4 x^2-1\right)\left(4 x^2+1\right)$
$=\left[(2 x)^2-(1)^2\right]\left(4 x^2+1\right)$
$=(2 x-1)(2 x+1)\left(4 x^2+1\right)$
$=\left(4 x^2\right)^2-(1)^2$
$=\left(4 x^2-1\right)\left(4 x^2+1\right)$
$=\left[(2 x)^2-(1)^2\right]\left(4 x^2+1\right)$
$=(2 x-1)(2 x+1)\left(4 x^2+1\right)$
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