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How do you factor the expression $a^{4} - 81$ $a^4 - 81$ ?

1 Answer

Dec 9, 2015

$a^{4} - 81 = (a^{2} + 9) (a + 3) (a - 3)$ $a^4-81=(a^2+9)(a+3)(a-3)$

Explanation:

$(a^{4} - 81) = ({(a^{2})}^{2} - 9^{2})$ $(a^4-81)=((a^2)^2-9^2)$
and is therefore the difference of squares with factors
$XXX (a^{2} + 9) (a^{2} - 9)$ $color(white)("XXX")(a^2+9)(a^2-9)$

but $(a^{2} - 9) = (a^{2} - 3^{2})$ $(a^2-9) = (a^2-3^2)$
is also the difference of squares with factors
$XXX (a + 3) (a - 3)$ $color(white)("XXX")(a+3)(a-3)$

$---------------------------------------------------------------------$ $"---------------------------------------------------------------------"$

Remember: Factoring the Difference of Squares
$XXX (x^{2} - y^{2}) = (x + y) (x - y)$ $color(white)("XXX")(x^2-y^2) = (x+y)(x-y)$

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