How do you factor #y= x^3 + 3x^2 + 9x + 27#?

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Answer 1 · 2018-05-05T07:03:57

#y = (x^2+9)(x+3)#

#color(white)(y) = (x-3i)(x+3i)(x+3)#

Given:

#y = x^3+3x^2+9x+27#

Note that the ratio of the first and second terms is the same as that of the third and fourth terms.

So this cubic will factor by grouping:

#y = x^3+3x^2+9x+27#

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

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

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

The remaining quadratic #x^2+9# is positive for all real values of #x#, so has no linear factors with real coefficients.

It can be factored with complex coefficients as a difference of squares...

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

where #i# is the imaginary unit, satisfying #i^2=-1#