Master Polynomial Concepts with Comprehensive Resources

When the speed of a moving object is doubled, its
A) acceleration is doubled
B) weight is doubled
C) K.E. is doubled
D) K.E. becomes 4 times

Find the zeroes of the quadratic polynomial $4x^2-4x-3$ and verify the relationship between the zeroes and the coefficients of the polynomial.

Find the sum of the first 100 natural numbers.

The highest exponent of a variable in a polynomial is called degree of the polynomial. Is this statement true or false?
True of False?

$\frac{x^2{y}^{-4}{z}^3\times x^{-5}{y}^5{z}^{-6}}{x^6{y}^2{z}^6\times x^{-4}{y}^{17}\times x^{13}{z}^{15}}\times x^2{y}^{-3}$

How to find the roots for $x^3+2x+2$

A sequence is defined by the recursive function f(n+1)=f(n)−2. If f(1)=10 , what is f(3)?

Infinite series representation for root of polynomials?
Given a polynomial $p(x)=a_n{x}^n+\cdots <!--\; \cdots \; -->+a_{1}x+a_0$, can every root of the polynomial be represented as $\underset{k=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{b}_k$ with the $b_k$'s being a function of $a_0,\dots <!--\; \dots \; -->,a_n$ using only elementary operations of arithmetic and taking roots?

Finding the integral closure of k[x] in $k(x)(\sqrt{f})$, where $f(x)=x^6+t{x}^5+t^2{x}^3+t\in <!--\; \in \; -->k[x]$
Let F be a field of characteristic 2. Let $k:=F(t)$, the field of rational functions in the variable t. Let $f(x)=x^6+t{x}^5+t^2{x}^3+t\in <!--\; \in \; -->k[x]$.
We want to show that the integral closure of k[x] in $k(x)(\sqrt{f})$ is $k[x][\sqrt{f}]$.
So let $\mathrm{\alpha <!--\; \alpha \; -->}=m+n\sqrt{f}\in <!--\; \in \; -->k(x)(\sqrt{f})$, where $m,n\in <!--\; \in \; -->k(x)$. Since F(t) is a field, k[x] is a PID and therefore $\mathrm{\alpha <!--\; \alpha \; -->}$ is integral over k[x] if and only if its minimal polynomial in k(x) has coefficients in k[x]. Let's then find the minimal polynomial of $\mathrm{\alpha <!--\; \alpha \; -->}$.
$\begin{array}{rl}p(y)& =(y-<!--\; -\; -->(m+n\sqrt{f}))(y-<!--\; -\; -->(m-<!--\; -\; -->n\sqrt{f}))\\ & =y^2-<!--\; -\; -->2my+(m^2-<!--\; -\; -->n^{2}f)\end{array}$
and since $m\in <!--\; \in \; -->k(x)=F(t)k(x)$ and the characteristic of F equals 2 we get that $2my=0$. Hence
$p(y)=y^2+(m^2-<!--\; -\; -->n^{2}f)$
Thus, $\mathrm{\alpha <!--\; \alpha \; -->}=m+n\sqrt{f}$ is integral over k[x] iff $m^2-<!--\; -\; -->n^{2}f\in <!--\; \in \; -->k[x]$. I' m pretty sure that I'm on the right course but I don' t know how to continue.

Give an example of a cubic polynomial

Does two's complement arithmetic produce a field isomorphic to $GF(2^n$?
From what I understand, we have these two isomorphisms:
(TC,+) is isomorphic to the cyclic group $\mathbb{Z}/2^n\mathbb{Z}$.
(TC,∗) is isomorphic to the multiplicative group of polynomials.
If this is correct, can we conclude that two's complement arithmetic produces a finite field isomorphic to $GF(2^n$?
If not, what algebraic structure, if any, does two's complement representation and arithmetic produce? Because there just seems to be something there.

Polynomial problem involving divisibility, prime numbers, monotony
Let f be a polynomial function, with integer coefficients, strictly increasing on $\mathbb{N}$ such that $f(0)=1$. Show that it doesn't exist any arithmetic progression of natural numbers with ratio $r>0$ such that the value of function f in every term of the progression is a prime number. I believe that the solution includes a reductio ad absurdum, but I don't know how to solve it.

Modular operation on polynomial rings
I originally started modular arithmetic by the following:
1mod2
1/2 is 0.5
0 times 2 is 0
$1-<!--\; -\; -->0=1$
1 equals 1mod2.
Is it the same way to compute a quotient of a polynomial ring such as ${\displaystyle \frac{\mathbb{C}[x_1,\dots <!--\; \dots \; -->,x_n]}{x^2+y^2-<!--\; -\; -->z^2}}$
$x^3+2x{y}^2-<!--\; -\; -->2x{z}^2+xmod{x}^2+y^2-<!--\; -\; -->z^2$
${\displaystyle \frac{x^3+2x{y}^2-<!--\; -\; -->2x{z}^2+x}{x^2+y^2-<!--\; -\; -->z^2}}$
${\displaystyle \frac{(x^3+2x{y}^2-<!--\; -\; -->2x{z}^2+x)\times <!--\; \times \; -->(x^3+2x{y}^2-<!--\; -\; -->2x{z}^2+x)}{(x^2+y^2-<!--\; -\; -->z^2)}}$
$x^3+2x{y}^2-<!--\; -\; -->2x{z}^2+x-<!--\; -\; -->{\displaystyle \frac{(x^3+2x{y}^2-<!--\; -\; -->2x{z}^2+x)\times <!--\; \times \; -->(x^3+2x{y}^2-<!--\; -\; -->2x{z}^2+x)}{x^2+y^2-<!--\; -\; -->z^2}}$

Converting a polynomial ring to a numerical ring (transport of structure)
One may be curious why one wishes to convert a polynomial ring to a numerical ring. But as one of the most natural number system is integers, and many properties of rings can be easily understood in parallel to ring of integers, I think converting a polynomial ring to a numerical (i.e. integer) ring is useful.
What I mean by converting to a numerical ring is: in the standard ring of integers, + and ⋅ are defined as in usual arithmetic. But is there universal way of converting any polynomial/monomial rings such that each object in the ring gets converted to an integer, and + and ⋅ can be defined differently from standard integer + and ⋅? This definition would be based on integer arithmetic, though.

Affine curve of an absolutely irreducible polynomial
Let $f\in <!--\; \in \; -->{\mathbb{F}}_q[X,Y]$ be an absolutely irreducible polynomial of degree n. Denote the affine curve defined by the equation $f(X,Y)=0$ over ${\mathbb{F}}_q$ by $\mathrm{\Gamma <!--\; \Gamma \; -->}({\mathbb{F}}_q)$, and let $d=\mathrm{deg}<!--\; \; -->\mathrm{\Gamma <!--\; \Gamma \; -->}$. The for each m there exists $q_0$ such that $|\mathrm{\Gamma <!--\; \Gamma \; -->}({\mathbb{F}}_q)|\ge <!--\; \ge \; -->m$ for all $q\ge <!--\; \ge \; -->q_0$.

Showing property for the derivative ${\mathrm{\partial <!--\; \partial \; -->}}_{x}T$ of a trigonometric polynomial
Let be
$T=\underset{n=0}{\overset{N}{\sum <!--\; \sum \; -->}}\stackrel{^<!--\; ^\; -->}{T}(n)e^{inx}$
a trigonometric polynomial of grade N without negative frequencies.
I wanna show that
${\mathrm{\partial <!--\; \partial \; -->}}_{x}T=-<!--\; -\; -->iN(F_N\ast <!--\; \ast \; -->T-<!--\; -\; -->T)$
Where $F_N\ast <!--\; \ast \; -->T$ meas the convolution of the Fejer Kernel and T.
might be easy..but I just can't work out the right conversion for this property..
SO
${\mathrm{\partial <!--\; \partial \; -->}}_{x}T=\underset{n=0}{\overset{N}{\sum <!--\; \sum \; -->}}\stackrel{^<!--\; ^\; -->}{T}(n)in{e}^{inx}$
$=\underset{n=0}{\overset{N}{\sum <!--\; \sum \; -->}}(\frac{1}{2\mathrm{\pi <!--\; \pi \; -->}}{\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->}}^{\mathrm{\pi <!--\; \pi \; -->}}T(y)e^{-<!--\; -\; -->iny}dy)e^{inx}in$
$=\frac{1}{2\mathrm{\pi <!--\; \pi \; -->}}{\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->}}^{\mathrm{\pi <!--\; \pi \; -->}}T(y)(\underset{n=0}{\overset{N}{\sum <!--\; \sum \; -->}}{e}^{in(x-<!--\; -\; -->y)})in$
from there on I get carried away in the wrong direction. is the derivative right?
Also..the Fejer Kernel can be expressed as the mean arithmetic value of the dirichtlet kernel so:
$F_N=\frac{1}{n+1}\underset{k=0}{\overset{n}{\sum <!--\; \sum \; -->}}{D}_k(x)$
Where $D_k=\underset{n=-<!--\; -\; -->k}{\overset{k}{\sum <!--\; \sum \; -->}}{e}^{inx}$ is the Dirichtlet Kernel

Let $f(x)=a{x}^3+b{x}^2+cx+d$, be a polynomial function, find relation between a,b,c,d such that it's roots are in an arithmetic/geometric progression. (separate relations)
So for the arithmetic progression I took let $\mathrm{\alpha <!--\; \alpha \; -->}=x_2$ and r be the ratio of the arithmetic progression.
We have:
$x_1=\mathrm{\alpha <!--\; \alpha \; -->}-<!--\; -\; -->2r,x_2=\mathrm{\alpha <!--\; \alpha \; -->},x_3=\mathrm{\alpha <!--\; \alpha \; -->}+2r$
Therefore:
$x_1+x_2+x_3=-<!--\; -\; -->\frac{b}{a}=3\mathrm{\alpha <!--\; \alpha \; -->}$
$x_1^2+x_2^2+x_3^2=9{\mathrm{\alpha <!--\; \alpha \; -->}}^2-<!--\; -\; -->2\frac{c}{a}\to <!--\; \to \; -->4r^2=\frac{b^2-<!--\; -\; -->3ac}{3a^2}$
$x_1{x}_2{x}_3=\mathrm{\alpha <!--\; \alpha \; -->}({\mathrm{\alpha <!--\; \alpha \; -->}}^2-<!--\; -\; -->4r^2)=-<!--\; -\; -->\frac{d}{a}$
and we get the final result $2b^3+27a^{2}d-<!--\; -\; -->9abc=0$.
How should I take the ratio at the geometric progression for roots?
I tried something like
$x_1=\frac{\mathrm{\alpha <!--\; \alpha \; -->}}{q},x_2=\mathrm{\alpha <!--\; \alpha \; -->},x_3=\mathrm{\alpha <!--\; \alpha \; -->}q$
To get $x_1{x}_2{x}_3={\mathrm{\alpha <!--\; \alpha \; -->}}^3$ but it doesn't really work out..
Note:
I have to choose from this set of answers:
$\text{(a)}{a}^{2}b=c^{2}d\text{(b)}{a}^2{b}^2=c^{2}d\text{(c)}a{b}^3=c^{3}d$
$\text{(d)}a{c}^3=b^{3}d\text{(e)}ac=bd\text{(f)}{a}^{3}c=b^{3}d$

Integral roots for $f(x)=41$ if $f(x)=37$ has 5 distinct integral roots.
Given a polynomial f(x) with integral coefficients and $f(x)=37$ has 5 distinct integral roots, find the number of integral roots of $f(x)=41$?
My Approach: Say $f(x)=(x-<!--\; -\; -->r_1)(x-<!--\; -\; -->r_2)(x-<!--\; -\; -->r_3)(x-<!--\; -\; -->r_4)(x-<!--\; -\; -->r_5)g(x)+37$, where ri are the distinct integers.
Now for $f(x)=41$ we have $(x-<!--\; -\; -->r_1)(x-<!--\; -\; -->r_2)(x-<!--\; -\; -->r_3)(x-<!--\; -\; -->r_4)(x-<!--\; -\; -->r_5)g(x)=4$, so the factors can be $\pm <!--\; \pm \; -->1,\pm <!--\; \pm \; -->2$ or $\pm <!--\; \pm \; -->4$. Given $r_i$ are distinct at most two of them will give $\pm <!--\; \pm \; -->1$, then there can be both of $\pm <!--\; \pm \; -->2$ or one of $\pm <!--\; \pm \; -->4$. This is where I get lost, since even if I use all of $\pm <!--\; \pm \; -->1,\pm <!--\; \pm \; -->2$, I will be still be left with one $x-<!--\; -\; -->r_i$ factor. What about that? Does it matter that 37 and 41 are primes, or is it just a coincidence?

Convolution product of arithmetic functions
An arithmetic function is a real-valued function whose domain is the set of positive integers. Define the convolution product of two arithmetic functions f and g to be the arithmetic function $f\star <!--\; \star \; -->g$, where
$(f\star <!--\; \star \; -->g)(n)=\underset{ij=n}{\sum <!--\; \sum \; -->}f(i)g(j),\text{ and }{f}^{\star <!--\; \star \; -->k}=f\star <!--\; \star \; -->f\star <!--\; \star \; -->\cdots <!--\; \cdots \; -->\star <!--\; \star \; -->f(k\text{ times}).$
We say that two arithmetic functions f and g are dependent if there exists a nontrivial polynomial of two variables $P(x,y)=\underset{i,j}{\sum <!--\; \sum \; -->}{a}_{ij}{x}^i{y}^j$ with real coefficients such that
$P(f,g)=\underset{i,j}{\sum <!--\; \sum \; -->}{a}_{ij}{f}^{\star <!--\; \star \; -->i}\star <!--\; \star \; -->g^{\star <!--\; \star \; -->j}=0$
and say that they are independent if they are not dependent. Let p and q be two distinct primes and set
$f_1(n)=\{\begin{array}{l}1\text{if }n=p,\\ 0\text{otherwise};\end{array}{f}_2(n)=\{\begin{array}{l}1\text{if }n=q,\\ 0\text{otherwise}.\end{array}$
Prove that $f_1$ and $f_2$ are independent.
My book says that
$f_1^{\star <!--\; \star \; -->i}\star <!--\; \star \; -->f_2^{\star <!--\; \star \; -->j}(n)=\{\begin{array}{l}1\text{if}n=p^i{q}^j\\ 0\text{otherwise}\end{array}$
and the result follows from that, but how do they get that and how does the result follow from that?

How many distinct degree 7 polynomials are there over the modular arithmeic modulo 7?
If it's infinite, is it countable or uncountable infinite?
I am a newbie to this topic... I don't know what modular arithmetic for polynomials means. Can someone please give me a link where I can learn?