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How do you factor completely: $16 x^{4} - 1$ $16x^4 - 1$ ?

1 Answer

Apr 25, 2018

$16 x^{4} - 1 = (4 x^{2} + 1) (2 x + 1) (2 x - 1)$ $16x^4-1=color(blue)((4x^2+1)(2x+1)(2x-1)$

Explanation:

Remember
$XXX (a^{2} - b^{2}) = (a + b) (a - b)$ $color(white)("XXX")(a^2-b^2)=(a+b)(a-b)$

$(16 x^{4} - 1)$ $(16x^4-1)$ is of this form (with $a = (4 x^{2})$ $a=(4x^2)$ and $b = 1$ $b=1$ ).

So
$XXX (16 x^{4} - 1) = (4 x^{2} + 1) (4 x^{2} - 1)$ $color(white)("XXX")(16x^4-1)=(4x^2+1)(4x^2-1)$

...but we note that $(4 x^{2} - 1)$ $(4x^2-1)$ is also of this form (with $a = 2 x$ $a=2x$ and $b = 1$ $b=1$ )
So $(4 x^{2} - 1) = (2 x + 1) (2 x - 1)$ $(4x^2-1)=(2x+1)(2x-1)$

and, completely factored we have
$XXX (16 x^{4} - 1) = (4 x^{2} + 1) (2 x + 1) (2 x - 1)$ $color(white)("XXX")(16x^4-1)=(4x^2+1)(2x+1)(2x-1)$

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