How do you factor the expression #3x - 9x^3#?

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Answer 1 · 2015-12-28T08:33:48

#3x-9x^3#

#= 3x(1-3x^2)#

#= 3x(1-sqrt(3)x)(1+sqrt(3)x)#

#= x(sqrt(3)-3x)(sqrt(3)+3x)#

We can use the difference of squares identity:

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

to factor with irrational coefficients.

First notice that both terms are divisible by #3x#, so separate that out as a factor:

#3x-9x^3#

#= 3x(1-3x^2)#

#= 3x(1^2-(sqrt(3)x)^2)#

#= 3x(1-sqrt(3)x)(1+sqrt(3)x)#

Or if you prefer, split that factor of #3# between the two binomials:

#3x-9x^3#

#= x(3-9x^2)#

#= x((sqrt(3))^2-(3x)^2)#

#= x(sqrt(3)-3x)(sqrt(3)+3x)#