I'd like to give my students a little visualization of the fact: $$x^n-a^n=(x-a)(x^{n-1}+x^{n-2}a+\cdots+x a^{n-2}+a^{n-1})$$

So, I've tried:

list = Table[PolynomialQuotient[x^n - a^n, x - a, x], {n, 2, 12}];
Column[list] // TraditionalForm

Which gave me the following output.

a^2+a x+x^2
a^3+a^2 x+a x^2+x^3
a^4+a^3 x+a^2 x^2+a x^3+x^4
a^5+a^4 x+a^3 x^2+a^2 x^3+a x^4+x^5
a^6+a^5 x+a^4 x^2+a^3 x^3+a^2 x^4+a x^5+x^6
a^7+a^6 x+a^5 x^2+a^4 x^3+a^3 x^4+a^2 x^5+a x^6+x^7
a^8+a^7 x+a^6 x^2+a^5 x^3+a^4 x^4+a^3 x^5+a^2 x^6+a x^7+x^8
a^9+a^8 x+a^7 x^2+a^6 x^3+a^5 x^4+a^4 x^5+a^3 x^6+a^2 x^7+a x^8+x^9
a^10+a^9 x+a^8 x^2+a^7 x^3+a^6 x^4+a^5 x^5+a^4 x^6+a^3 x^7+a^2 x^8+a x^9+x^10
a^11+a^10 x+a^9 x^2+a^8 x^3+a^7 x^4+a^6 x^5+a^5 x^6+a^4 x^7+a^3 x^8+a^2 x^9+a x^10+x^11

Is there a way I can put the x's before the a's, for example, arranging $$a+x\qquad\text{as}\qquad x+a,$$ and $$a^2+ax+x^2\qquad\text{as}\qquad x^2+xa+a^2,$$ etc.?

  • 1
    $\begingroup$ Some related Q&A: (33130), (37339) -- The solution to the second might work here. $\endgroup$
    – Michael E2
    Jan 8, 2016 at 19:14

1 Answer 1


Disclaimer: Doesn't work with TraditionalForm

list = Table[PolynomialQuotient[x^n - a^n, x - a, x], {n, 2, 12}];

ClearAttributes[{Plus, Times}, Orderless]

list /.
   {h_[a, x] :> h[x, a],
    t : Times[Power[a, i_], Power[x, j_]] :> Reverse@t,
    t : Times[a, Power[x, j_]] :> Reverse@t,
    t : Times[Power[a, i_], x] :> Reverse@t} /.
    Plus[a : Power[a, i_], b__, c : Power[x, j_]] :> Plus[c, b, a] // Column  

enter image description here

Don't forget

SetAttributes[{Plus, Times}, Orderless]

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