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?
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$\begingroup$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).
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