I am doing some self-assigned homework questions. One question asks me to factorize $x^4 + 27x$.

I used the factor theorem to solve it. I observed that $f(0) = 0$ and $f(-3) = 0$, so $x$ and $(x + 3)$ are factors. I used these factors to obtain the last factor: $\frac{x^4 + 27x}{x(x + 3)} = x^2 - 3x + 9$. Therefore, $x ^ 4 + 27x = x(x + 3)(x^2 -3x + 9)$.

I don't know if this is how I'm "supposed to" solve the problem, because I am doing this on my own. Is there a way to factorize $x^4 + 27x$ without employing the factor theorem?


There is no "best" and "worst" way, there are only "right" and "wrong" ways. (Yours is right.) Of course, for educational purposes, some may be better than others because they provide additional insights, and using factor theorem is certainly one of those.

One alternative that comes to my mind is that you may know the identity $a^3+b^3=(a+b)(a^2-ab+b^2)$, so you could've noticed that $x^4+27x=x(x^3+3^3)=x(x+3)(x^2-3x+9)$ straight away. This identity is BTW a special case of $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1})$ for $x=a, y=-b, n=3$, and is good to know it (and its special cases).

  • $\begingroup$ Is there a name for the $x^n - y^n = ...$ identity? $\endgroup$ – Flux Oct 22 '20 at 10:52
  • $\begingroup$ @Flux You'd probably just call it the difference of two $n$th powers. The $n=2$ case is called DOTS (difference of two squares); compare with SOTS (sum of two squares, $x^2+y^2=\prod_\pm(x\pm iy)$. $\endgroup$ – J.G. Oct 22 '20 at 10:56
  • $\begingroup$ Indeed, the factorization of $x^n-y^n$ is how we prove the remainder theorem. $\endgroup$ – J.G. Oct 22 '20 at 10:56
  • $\begingroup$ Yep, AoPS is calling it "difference of powers": artofproblemsolving.com/wiki/… . They also cover "sum of powers" for an odd exponent (like yours, $n=3$), which is indeed another special case of "difference of powers" with the sign of the second variable flipped. $\endgroup$ – Stinking Bishop Oct 22 '20 at 10:59

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