How do you factor #4x^3 + 36#?

1 Answer
George C.
May 9, 2016

#4x^3+36 = 4(x+root(3)(9))(x^2-root(3)(9)x+3root(3)(3))#

Explanation:

First separate out the common scalar factor #4#:

#4x^3+36 = 4(x^3+9)#

Note that #x^3# is a perfect cube, but #9# is not - at least not a cube of a rational number.

We can still use the sum of cubes identity:

#a^3+b^3=(a+b)(a^2-ab+b^2)#

with #a=x# and #b=root(3)(9)# as follows:

#x^3+9#

#=x^3+(root(3)(9))^3#

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

#=(x+root(3)(9))(x^2-root(3)(9)x+root(3)(81))#

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

So:

#4x^3+36 = 4(x+root(3)(9))(x^2-root(3)(9)x+3root(3)(3))#

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