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)#