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Michael Spivak, Calculus (4th edn 2008), p 13. I know this monograph is aimed at undergraduates (not middle schoolers), but this kind of multiple-part question resurfaces on standardized tests for 13 year olds (henceforth y.o.).

  1. Prove the following:

(v) $x^n - y^n = (x - y)(x^{n - 1} + x^{n - 2}y + \cdots + xy^{n - 2} + y{^n - 1}).$
(vi) $x^3 + y^3 = (x + y)(x^ 2- xy + y^2)$. (There is a particularly easy way to do this, using (iv), and it will show you how to find a factorization for $x^n + y^n$ $\color{peru}{\; whenever \; n \; is \; odd}.$)

Ibid, p 619.

(vi) Replace $y$ by $-y$ in (iv).

Answers that use Cyclotomic Polynomials, Modular Arithmetic, Complex Numbers are too Delphic for 13 y.o.!

Brian M. Scott’s answer is TOO gruff and sketchy for 13 y.o. to fathom. To wit, 13 y.o. know that $(a + b)[a^{n – 1} – a^{n – 2}b + . . . + \color{mediumspringgreen}{+} ab^{n – 2} \color{red}{-} b^{n – 1}] \equiv a^n – b^n$. But this doesn’t answer their question! How can 13 y.o. intuit why they can't factor out $(x + y)$ from $x^n + y^n$ for all $n \in \mathbb{N}$?

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    $\begingroup$ Maybe start with $y=1$ and show that: $\frac{x^2+1}{x+1}$ is not a polynomial. Do this by examining what happens as $x \to -1$. $\endgroup$ Commented Oct 15, 2023 at 1:31
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    $\begingroup$ Do you mean "for all natural numbers $n$"? $\endgroup$
    – mweiss
    Commented Oct 15, 2023 at 2:03
  • $\begingroup$ "How do we know that $c$ can't exist, such that $(x+y)c=x^n+y^n$?" with $n,x,y\in\mathbb R$. Try $c=1$, $n=1$. $\endgroup$
    – JRN
    Commented Oct 15, 2023 at 2:09
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    $\begingroup$ 3 hours old, and already 3 answers posted as comments! Post answers as answers instead, so they can be edited, voted on, and commented on. $\endgroup$ Commented Oct 15, 2023 at 4:22
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    $\begingroup$ "this kind of multiple-part question resurfaces on standardized tests for 13 year olds" Really? That does not seem at all reasonable. $\endgroup$
    – Sue VanHattum
    Commented Oct 15, 2023 at 18:17

3 Answers 3

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We can use the factor theorem, unless you think it’s also too “Delphic” for your students. The factor theorem tells us that a polynomial $p(x)$ has a factor $(x-r)$ if and only if $p(r)=0$. Equivalently, $p(x)$ has a factor $(x-r)$ if and only if $r$ is a root of $p(x)$.

Let’s consider the polynomial defined as $p(x) = x^n + a^n$, where $a$ is non-zero. It has a factor $(x+a)$ if and only if $p(-a)=0$. But $$p(-a)=(-a)^n+a^n= \begin{cases} 0 &\text{ if $n$ is odd,} \\ 2a^n &\text{if $n$ is even,} \end{cases}$$ therefore $(x+a)$ is a factor of $x^n+a^n$ if and only if $n$ is odd.

Furthermore, if $n$ is even, $x^n+a^n$ has no linear factors as a polynomial over the real numbers. This is because the roots of $x^n+a^n$ are the solutions to the equation $x^n=-a^n$. This equation has no real solutions.

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The question is framed very oddly, so I'm not really sure what you're looking for.

If the student understands that p(x) being factorable means that p(x)=0 has roots when any linear factor equals 0, then you can use desmos to graph y=𝑥^n + a^n. For all real nonzero values of a and even values of n, this does not cross the x-axis, hence there are no linear factors.

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Maybe show them $x^2+y^2$ as an example? It is unfactorable. Let them try.

If anything this is a lor easier than proving general quintic has no formula, which you can only assert to students not prove.

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    $\begingroup$ For $x^2+y^2$ I would add the following: if it were factorable, the set of solutions of $x^2+y^2 = 0$ would include a line, but it doesn't. This argument should be understandable to younger students. $\endgroup$ Commented Oct 15, 2023 at 19:21

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