# Can I factorize $x^4 + 27x$ without using the factor theorem?

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?

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).
• Is there a name for the $x^n - y^n = ...$ identity? – Flux Oct 22 '20 at 10:52
• @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, $x^2+y^2=\prod_\pm(x\pm iy)$. – J.G. Oct 22 '20 at 10:56
• Indeed, the factorization of $x^n-y^n$ is how we prove the remainder theorem. – J.G. Oct 22 '20 at 10:56
• 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. – Stinking Bishop Oct 22 '20 at 10:59