Showing a polynomial is divisible by another over Z_5

1. Oct 22, 2016

PsychonautQQ

1. The problem statement, all variables and given/known data
Show that x^7 + 3x^6 + 3x^5 - x^3 - 3x^2 - 3x is divisible by x^5-x

2. Relevant equations

3. The attempt at a solution
So i did polynomial long division and as a quotient so far I have x^2+3x, and it appears that my remainder is going to be 3(x^3-x). Does this mean that I did something wrong? I'm relooking the steps of my long division very carefully and I can't find any mistake. I mean, after doing the step involving subtraction I arrive at 3(x^3-x), and obvioulsy I can't use x^5-x to divide it anymore because it's of a greater degree. Anyone have any insight?

2. Oct 22, 2016

Staff: Mentor

I think so, I have another result. Your term of smallest degree is $3x$, but the the lowest degree of $(x^7 + 3x^6 + 3x^5 - x^3 - 3x^2 - 3x) \, : \, (x^5-x)$ should be $(-3x)\, : \, (-x) = 3$ if there is no remainder.
You can always check your calculations by multiplying back and see if you will get the entire and correct polynomial.
My advice is to do the division without abbreviations, i.e. for $p(x) \, : \, q(X)$
• highest term of $p(x)$ $:$ highest term of $q(x)$, say $\rightarrow r_1(x)$
• multiply $r_1(x) \,\cdot\, q(x)$, say $\rightarrow r'_1(x)$
• multiply $r'_1(x) \,\cdot\, q(x)$ by $(-1)$, say $\rightarrow r''_1(x)$
• add $p(x)$ and $r''_1(x)$, say $\rightarrow p_1(x)$
• start next step, i.e. highest term of $p_1(x)$ $:$ highest term of $q(x)$, say $\rightarrow r_2(x)$
• etc.
This method is somehow safer than to directly subtract terms, which is a potential cause for sign errors.
My guess is, that you simply have forgotten one or two terms of the long $p(x)$.

3. Oct 23, 2016

ehild

You certainly did something wrong during long division. And you can get the result much easier.
Both polynomials have the common factor x, so you can divide both of them by x. x^6 + 3x^5 + 3x^4
-x^2 - 3x - 3 is divisible by x^4-1.Try grouping the terms.

(x^6 + 3x^5 + 3x^4)-(x^2 + 3x + 3)=x4(x^2 + 3x + 3)-(x^2 + 3x + 3)

Last edited: Oct 23, 2016
4. Oct 23, 2016

pasmith

If it is true that $x^5 - x$ divides $x^7 + 3x^6 + 3x^5 - x^3 - 3x^2 - 3x$ then we must have $$x^7 + 3x^6 + 3x^5 - x^3 - 3x^2 - 3x = (x^5 - x)(x^2 + bx + 3)$$ for some $b$ which can be found by expanding the right hand side and comparing coefficients of powers of $x$. (The coefficients of $x^2$ and $x^0$ in the unknown factor are fixed so that $x^7 = x^5x^2$ and $-3x = (-x)(3)$.)

If comparing coefficients in this way leads to incompatible conditions which $b$ must satisfy then either our assumption of divisibility was false or we have a made an error in the algebra.

This argument for establishing divisibility is less prone to algebraic error than doing polynomial long division and checking the remainder.