How do you factor completely $5x^3 - 45x$ ?

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Answer 1 · 2016-04-26T21:07:21

$5x^3-45x = 5x(x-3)(x+3)$

Notice that both of the terms are divisible by $5x$ , so we can separate that out as a factor first, then use the difference of squares identity:

$a^2-b^2=(a-b)(a+b)$

with $a=x$ and $b=3$ as follows:

$5x^3-45x = 5x(x^2-9) = 5x(x^2-3^2) = 5x(x-3)(x+3)$

How do you factor completely 5x^3 - 45x5x^3 - 45x?