How do you factor $4x^5 - 9x^3$ ?

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Answer 1 · 2017-04-24T07:48:19

$= x^3(2x+3)(2x-3)$

Always take out a common factor first if there is one.

$4x^5-9x^3$

$= x^3(4x^2 -9)" "larr$ difference of squares

$= x^3(2x+3)(2x-3)$

To recognise Difference of squares:

$x^2 =y^2 = (x+y)(x-y)$

Answer 2 · 2017-04-24T08:00:58

$x^3(2x+3)(2x-3)$

$4x^5-9x^3$

Factor out $x^3$

$x^3(4x^2-9)$

Here we realize that both $4x^2$ and $9$ are perfect squares. So, we factor them as.

$color(red)((sqrtA+sqrtB) (sqrtA-sqrtB))$
I this case $color(red)A=4x^2$ and $color(red)B=9$

So,
$x^3(sqrt(4x^2)+sqrt9)(sqrt(4x^2)-sqrt9)$

Complete the square roots to get your final answer

$x^3(2x+3)(2x-3)$

How do you factor 4x^5 - 9x^3 4x^5 - 9x^3?