How do you factor x^3+27?

2 Answers
Anjali G
Nov 3, 2016

Because factoring x^3+27 is the same as finding where the graph passes through the x axis, we can just set the equation equal to zero and solve. #f(x) = (x+3)(x^2-3x+9)#

Explanation:

Let f(x) = x³ + 27
0 = x³ + 27
x³= -27
x = -3
This means that x = -3 is the only zero of the graph of f(x). Since we know #(x+3)# is one factor of #f(x)#, to find the 2nd factor,
#(x^3+27)/(x+3)#
We get the second factor to be #x^2-3x+9#.

Therefore, #f(x) = (x+3)(x^2-3x+9)#

George C.
Nov 3, 2016

#x^3+27 = (x+3)(x^2-3x+9)#

Explanation:

The sum of cubes identity can be written:

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

Note that both #x^3# and #27=3^3# are perfect cubes, so we can use the sum of cubes identity with #a=x# and #b=3# as follows:

#x^3+27 = x^3+3^3#

#color(white)(x^3+27) = (x+3)(x^2-3x+3^2)#

#color(white)(x^3+27) = (x+3)(x^2-3x+9)#

The remaining quadratic has no simpler linear factors with Real coefficients.

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