Showing a polynomial is divisible by another over Z_5

  • Thread starter PsychonautQQ
  • Start date
  • Tags
    Polynomial
In summary: In short, this is a safer method.In summary, if x^5 - x divides x^7 + 3x^6 + 3x^5 - x^3 - 3x^2 - 3x then x^7 + 3x^6 + 3x^5 - x^3 - 3x^2 - 3x can be reduced to (x^5 - x)(x^2 + bx + 3). Homework Statement Show that x^7 + 3x^6 + 3x^5 - x^3 - 3x^2 - 3x is divisible by x^5-x
  • #1
PsychonautQQ
784
10

Homework Statement


Show that x^7 + 3x^6 + 3x^5 - x^3 - 3x^2 - 3x is divisible by x^5-x

Homework Equations

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?
 
Physics news on Phys.org
  • #2
PsychonautQQ said:

Homework Statement


Show that x^7 + 3x^6 + 3x^5 - x^3 - 3x^2 - 3x is divisible by x^5-x

Homework Equations

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 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.
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?
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)##.
 
  • Like
Likes PsychonautQQ
  • #3
PsychonautQQ said:

Homework Statement


Show that x^7 + 3x^6 + 3x^5 - x^3 - 3x^2 - 3x is divisible by x^5-x
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:
  • Like
Likes PsychonautQQ
  • #4
PsychonautQQ said:

Homework Statement


Show that x^7 + 3x^6 + 3x^5 - x^3 - 3x^2 - 3x is divisible by x^5-x

Homework Equations

The Attempt at a Solution


So i did polynomial long division

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

If comparing coefficients in this way leads to incompatible conditions which [itex]b[/itex] 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.
 
  • Like
Likes PsychonautQQ

FAQ: Showing a polynomial is divisible by another over Z_5

1. How do you determine if a polynomial is divisible by another over Z_5?

To determine if a polynomial is divisible by another over Z_5, you can use the long division method or the Euclidean algorithm. Both methods involve dividing the polynomial by the other and checking if the remainder is equal to zero.

2. What is the significance of Z_5 in this context?

Z_5, also known as the integers modulo 5, is a mathematical structure that is used to represent a finite set of numbers from 0 to 4. In this context, Z_5 is used because it allows us to perform operations on polynomials with coefficients from 0 to 4, which is necessary for determining divisibility.

3. Can you show an example of proving divisibility using Z_5?

Yes, for example, let's say we want to show that the polynomial x^3 + 2x^2 + x is divisible by the polynomial x + 4 over Z_5. Using the long division method, we get x^2 + 3x + 2 as the quotient and 1 as the remainder. Since the remainder is not equal to zero, we can conclude that x^3 + 2x^2 + x is not divisible by x + 4 over Z_5.

4. Are there any other methods for proving divisibility over Z_5?

Yes, another method is using the fact that if a polynomial is divisible by another over Z_5, all of its coefficients must be divisible by the coefficients of the other polynomial. This method is useful when the polynomials have higher degrees and long division becomes too tedious.

5. Can divisibility over Z_5 be extended to other mathematical structures?

Yes, divisibility can be extended to other structures such as Z_n, where n is any positive integer. However, the process and methods for proving divisibility may vary depending on the structure. For example, in Z_p, where p is a prime number, the long division method is not applicable and the use of modular arithmetic is required.

Back
Top