Recent questions in Polynomial arithmetic

Algebra IIAnswered question

Paloma Owens 2023-03-20

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

A) acceleration is doubled

B) weight is doubled

C) K.E. is doubled

D) K.E. becomes 4 times

Algebra IIAnswered question

nestalno4szl 2023-02-20

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

Algebra IIAnswered question

Deven Blanchard 2023-01-10

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

True of False?

True of False?

Algebra IIAnswered question

Gael Woodward 2022-12-26

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

Algebra IIAnswered question

Milagros Moon 2022-11-27

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

Algebra IIAnswered question

aplaya4lyfeSS1 2022-11-25

Infinite series representation for root of polynomials?

Given a polynomial $p(x)={a}_{n}{x}^{n}+\cdots +{a}_{1}x+{a}_{0}$, can every root of the polynomial be represented as $\sum _{k=0}^{\mathrm{\infty}}{b}_{k}$ with the ${b}_{k}$'s being a function of ${a}_{0},\dots ,{a}_{n}$ using only elementary operations of arithmetic and taking roots?

Given a polynomial $p(x)={a}_{n}{x}^{n}+\cdots +{a}_{1}x+{a}_{0}$, can every root of the polynomial be represented as $\sum _{k=0}^{\mathrm{\infty}}{b}_{k}$ with the ${b}_{k}$'s being a function of ${a}_{0},\dots ,{a}_{n}$ using only elementary operations of arithmetic and taking roots?

Algebra IIAnswered question

blogmarxisteFAu 2022-11-24

Finding the integral closure of k[x] in $k(x)(\sqrt{f})$, where $f(x)={x}^{6}+t{x}^{5}+{t}^{2}{x}^{3}+t\in k[x]$

Let F be a field of characteristic 2. Let $k:=F(t)$, the field of rational functions in the variable t. Let $f(x)={x}^{6}+t{x}^{5}+{t}^{2}{x}^{3}+t\in k[x]$.

We want to show that the integral closure of k[x] in $k(x)(\sqrt{f})$ is $k[x][\sqrt{f}]$.

So let $\alpha =m+n\sqrt{f}\in k(x)(\sqrt{f})$, where $m,n\in k(x)$. Since F(t) is a field, k[x] is a PID and therefore $\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 $\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 $m\in k(x)=F(t)k(x)$ and the characteristic of F equals 2 we get that $2my=0$. Hence

$p(y)={y}^{2}+({m}^{2}-{n}^{2}f)$

Thus, $\alpha =m+n\sqrt{f}$ is integral over k[x] iff ${m}^{2}-{n}^{2}f\in k[x]$. I' m pretty sure that I'm on the right course but I don' t know how to continue.

Let F be a field of characteristic 2. Let $k:=F(t)$, the field of rational functions in the variable t. Let $f(x)={x}^{6}+t{x}^{5}+{t}^{2}{x}^{3}+t\in k[x]$.

We want to show that the integral closure of k[x] in $k(x)(\sqrt{f})$ is $k[x][\sqrt{f}]$.

So let $\alpha =m+n\sqrt{f}\in k(x)(\sqrt{f})$, where $m,n\in k(x)$. Since F(t) is a field, k[x] is a PID and therefore $\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 $\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 $m\in k(x)=F(t)k(x)$ and the characteristic of F equals 2 we get that $2my=0$. Hence

$p(y)={y}^{2}+({m}^{2}-{n}^{2}f)$

Thus, $\alpha =m+n\sqrt{f}$ is integral over k[x] iff ${m}^{2}-{n}^{2}f\in k[x]$. I' m pretty sure that I'm on the right course but I don' t know how to continue.

Algebra IIAnswered question

Emmanuel Giles 2022-11-23

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.

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.

Algebra IIAnswered question

Anton Huynh 2022-11-23

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.

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.

Algebra IIAnswered question

Zackary Diaz 2022-11-23

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 $\frac{\mathbb{C}[{x}_{1},\dots ,{x}_{n}]}{{x}^{2}+{y}^{2}-{z}^{2}}$

${x}^{3}+2x{y}^{2}-2x{z}^{2}+xmod\phantom{\rule{thickmathspace}{0ex}}{x}^{2}+{y}^{2}-{z}^{2}$

$\frac{{x}^{3}+2x{y}^{2}-2x{z}^{2}+x}{{x}^{2}+{y}^{2}-{z}^{2}}$

$\frac{({x}^{3}+2x{y}^{2}-2x{z}^{2}+x)\times ({x}^{3}+2x{y}^{2}-2x{z}^{2}+x)}{({x}^{2}+{y}^{2}-{z}^{2})}$

${x}^{3}+2x{y}^{2}-2x{z}^{2}+x-{\displaystyle \frac{({x}^{3}+2x{y}^{2}-2x{z}^{2}+x)\times ({x}^{3}+2x{y}^{2}-2x{z}^{2}+x)}{{x}^{2}+{y}^{2}-{z}^{2}}}$

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 $\frac{\mathbb{C}[{x}_{1},\dots ,{x}_{n}]}{{x}^{2}+{y}^{2}-{z}^{2}}$

${x}^{3}+2x{y}^{2}-2x{z}^{2}+xmod\phantom{\rule{thickmathspace}{0ex}}{x}^{2}+{y}^{2}-{z}^{2}$

$\frac{{x}^{3}+2x{y}^{2}-2x{z}^{2}+x}{{x}^{2}+{y}^{2}-{z}^{2}}$

$\frac{({x}^{3}+2x{y}^{2}-2x{z}^{2}+x)\times ({x}^{3}+2x{y}^{2}-2x{z}^{2}+x)}{({x}^{2}+{y}^{2}-{z}^{2})}$

${x}^{3}+2x{y}^{2}-2x{z}^{2}+x-{\displaystyle \frac{({x}^{3}+2x{y}^{2}-2x{z}^{2}+x)\times ({x}^{3}+2x{y}^{2}-2x{z}^{2}+x)}{{x}^{2}+{y}^{2}-{z}^{2}}}$

Algebra IIAnswered question

Celeste Barajas 2022-11-23

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.

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.

Algebra IIAnswered question

Howard Nelson 2022-11-22

Affine curve of an absolutely irreducible polynomial

Let $f\in {\mathbb{F}}_{q}[X,Y]$ 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 $\mathrm{\Gamma}({\mathbb{F}}_{q})$, and let $d=\mathrm{deg}\mathrm{\Gamma}$. The for each m there exists ${q}_{0}$ such that $|\mathrm{\Gamma}({\mathbb{F}}_{q})|\ge m$ for all $q\ge {q}_{0}$.

Let $f\in {\mathbb{F}}_{q}[X,Y]$ 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 $\mathrm{\Gamma}({\mathbb{F}}_{q})$, and let $d=\mathrm{deg}\mathrm{\Gamma}$. The for each m there exists ${q}_{0}$ such that $|\mathrm{\Gamma}({\mathbb{F}}_{q})|\ge m$ for all $q\ge {q}_{0}$.

Algebra IIAnswered question

Arendrogfkl 2022-11-21

Showing property for the derivative ${\mathrm{\partial}}_{x}T$ of a trigonometric polynomial

Let be

$T=\sum _{n=0}^{N}\hat{T}(n){e}^{inx}$

a trigonometric polynomial of grade N without negative frequencies.

I wanna show that

${\mathrm{\partial}}_{x}T=-iN({F}_{N}\ast T-T)$

Where ${F}_{N}\ast T$ 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

${\mathrm{\partial}}_{x}T=\sum _{n=0}^{N}\hat{T}(n)in{e}^{inx}$

$=\sum _{n=0}^{N}(\frac{1}{2\pi}{\int}_{-\pi}^{\pi}T(y){e}^{-iny}dy){e}^{inx}in$

$=\frac{1}{2\pi}{\int}_{-\pi}^{\pi}T(y)(\sum _{n=0}^{N}{e}^{in(x-y)})in$

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:

${F}_{N}=\frac{1}{n+1}\sum _{k=0}^{n}{D}_{k}(x)$

Where ${D}_{k}=\sum _{n=-k}^{k}{e}^{inx}$ is the Dirichtlet Kernel

Let be

$T=\sum _{n=0}^{N}\hat{T}(n){e}^{inx}$

a trigonometric polynomial of grade N without negative frequencies.

I wanna show that

${\mathrm{\partial}}_{x}T=-iN({F}_{N}\ast T-T)$

Where ${F}_{N}\ast T$ 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

${\mathrm{\partial}}_{x}T=\sum _{n=0}^{N}\hat{T}(n)in{e}^{inx}$

$=\sum _{n=0}^{N}(\frac{1}{2\pi}{\int}_{-\pi}^{\pi}T(y){e}^{-iny}dy){e}^{inx}in$

$=\frac{1}{2\pi}{\int}_{-\pi}^{\pi}T(y)(\sum _{n=0}^{N}{e}^{in(x-y)})in$

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:

${F}_{N}=\frac{1}{n+1}\sum _{k=0}^{n}{D}_{k}(x)$

Where ${D}_{k}=\sum _{n=-k}^{k}{e}^{inx}$ is the Dirichtlet Kernel

Algebra IIAnswered question

Jefferson Booth 2022-11-21

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 $\alpha ={x}_{2}$ and r be the ratio of the arithmetic progression.

We have:

${x}_{1}=\alpha -2r,\phantom{\rule{1em}{0ex}}{x}_{2}=\alpha ,\phantom{\rule{1em}{0ex}}{x}_{3}=\alpha +2r$

Therefore:

${x}_{1}+{x}_{2}+{x}_{3}=-\frac{b}{a}=3\alpha $

${x}_{1}^{2}+{x}_{2}^{2}+{x}_{3}^{2}=9{\alpha}^{2}-2\frac{c}{a}\to 4{r}^{2}=\frac{{b}^{2}-3ac}{3{a}^{2}}$

${x}_{1}{x}_{2}{x}_{3}=\alpha ({\alpha}^{2}-4{r}^{2})=-\frac{d}{a}$

and we get the final result $2{b}^{3}+27{a}^{2}d-9abc=0$.

How should I take the ratio at the geometric progression for roots?

I tried something like

${x}_{1}=\frac{\alpha}{q},\phantom{\rule{1em}{0ex}}{x}_{2}=\alpha ,\phantom{\rule{1em}{0ex}}{x}_{3}=\alpha q$

To get ${x}_{1}{x}_{2}{x}_{3}={\alpha}^{3}$ but it doesn't really work out..

Note:

I have to choose from this set of answers:

$\text{(a)}\text{}{a}^{2}b={c}^{2}d\phantom{\rule{1em}{0ex}}\text{(b)}\text{}{a}^{2}{b}^{2}={c}^{2}d\phantom{\rule{1em}{0ex}}\text{(c)}\text{}a{b}^{3}={c}^{3}d$

$\text{(d)}\text{}a{c}^{3}={b}^{3}d\phantom{\rule{1em}{0ex}}\text{(e)}\text{}ac=bd\phantom{\rule{1em}{0ex}}\text{(f)}\text{}{a}^{3}c={b}^{3}d$

So for the arithmetic progression I took let $\alpha ={x}_{2}$ and r be the ratio of the arithmetic progression.

We have:

${x}_{1}=\alpha -2r,\phantom{\rule{1em}{0ex}}{x}_{2}=\alpha ,\phantom{\rule{1em}{0ex}}{x}_{3}=\alpha +2r$

Therefore:

${x}_{1}+{x}_{2}+{x}_{3}=-\frac{b}{a}=3\alpha $

${x}_{1}^{2}+{x}_{2}^{2}+{x}_{3}^{2}=9{\alpha}^{2}-2\frac{c}{a}\to 4{r}^{2}=\frac{{b}^{2}-3ac}{3{a}^{2}}$

${x}_{1}{x}_{2}{x}_{3}=\alpha ({\alpha}^{2}-4{r}^{2})=-\frac{d}{a}$

and we get the final result $2{b}^{3}+27{a}^{2}d-9abc=0$.

How should I take the ratio at the geometric progression for roots?

I tried something like

${x}_{1}=\frac{\alpha}{q},\phantom{\rule{1em}{0ex}}{x}_{2}=\alpha ,\phantom{\rule{1em}{0ex}}{x}_{3}=\alpha q$

To get ${x}_{1}{x}_{2}{x}_{3}={\alpha}^{3}$ but it doesn't really work out..

Note:

I have to choose from this set of answers:

$\text{(a)}\text{}{a}^{2}b={c}^{2}d\phantom{\rule{1em}{0ex}}\text{(b)}\text{}{a}^{2}{b}^{2}={c}^{2}d\phantom{\rule{1em}{0ex}}\text{(c)}\text{}a{b}^{3}={c}^{3}d$

$\text{(d)}\text{}a{c}^{3}={b}^{3}d\phantom{\rule{1em}{0ex}}\text{(e)}\text{}ac=bd\phantom{\rule{1em}{0ex}}\text{(f)}\text{}{a}^{3}c={b}^{3}d$

Algebra IIAnswered question

Jefferson Booth 2022-11-20

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 $\pm 1,\pm 2$ or $\pm 4$. Given ${r}_{i}$ are distinct at most two of them will give $\pm 1$, then there can be both of $\pm 2$ or one of $\pm 4$. This is where I get lost, since even if I use all of $\pm 1,\pm 2$, 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?

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 $\pm 1,\pm 2$ or $\pm 4$. Given ${r}_{i}$ are distinct at most two of them will give $\pm 1$, then there can be both of $\pm 2$ or one of $\pm 4$. This is where I get lost, since even if I use all of $\pm 1,\pm 2$, 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?

Algebra IIAnswered question

apopiw83 2022-11-20

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 $f\star g$, where

$(f\star g)(n)=\sum _{ij=n}f(i)g(j),\text{and}{f}^{\star k}=f\star f\star \cdots \star f\text{}(k\text{times}).$

We say that two arithmetic functions f and g are dependent if there exists a nontrivial polynomial of two variables $P(x,y)=\sum _{i,j}{a}_{ij}{x}^{i}{y}^{j}$ with real coefficients such that

$P(f,g)=\sum _{i,j}{a}_{ij}{f}^{\star i}\star {g}^{\star j}=0$

and say that they are independent if they are not dependent. Let p and q be two distinct primes and set

${f}_{1}(n)=\{\begin{array}{l}1\phantom{\rule{1em}{0ex}}\text{if}n=p,\\ 0\phantom{\rule{1em}{0ex}}\text{otherwise};\end{array}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}{f}_{2}(n)=\{\begin{array}{l}1\phantom{\rule{1em}{0ex}}\text{if}n=q,\\ 0\phantom{\rule{1em}{0ex}}\text{otherwise}.\end{array}$

Prove that ${f}_{1}$ and ${f}_{2}$ are independent.

My book says that

${f}_{1}^{\star i}\star {f}_{2}^{\star j}(n)=\{\begin{array}{l}1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{if}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}n={p}^{i}{q}^{j}\\ 0\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{otherwise}\end{array}$

and the result follows from that, but how do they get that and how does the result follow from that?

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 $f\star g$, where

$(f\star g)(n)=\sum _{ij=n}f(i)g(j),\text{and}{f}^{\star k}=f\star f\star \cdots \star f\text{}(k\text{times}).$

We say that two arithmetic functions f and g are dependent if there exists a nontrivial polynomial of two variables $P(x,y)=\sum _{i,j}{a}_{ij}{x}^{i}{y}^{j}$ with real coefficients such that

$P(f,g)=\sum _{i,j}{a}_{ij}{f}^{\star i}\star {g}^{\star j}=0$

and say that they are independent if they are not dependent. Let p and q be two distinct primes and set

${f}_{1}(n)=\{\begin{array}{l}1\phantom{\rule{1em}{0ex}}\text{if}n=p,\\ 0\phantom{\rule{1em}{0ex}}\text{otherwise};\end{array}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}{f}_{2}(n)=\{\begin{array}{l}1\phantom{\rule{1em}{0ex}}\text{if}n=q,\\ 0\phantom{\rule{1em}{0ex}}\text{otherwise}.\end{array}$

Prove that ${f}_{1}$ and ${f}_{2}$ are independent.

My book says that

${f}_{1}^{\star i}\star {f}_{2}^{\star j}(n)=\{\begin{array}{l}1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{if}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}n={p}^{i}{q}^{j}\\ 0\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{otherwise}\end{array}$

and the result follows from that, but how do they get that and how does the result follow from that?

Algebra IIAnswered question

django0a6 2022-11-20

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?

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?

If you have been looking for polynomial math problems and answers but had no luck, we are happy to provide you with both questions and answers to various polynomial math problems. These are used for the description and prediction of traffic patterns, yet some equations that implement polynomials are also encountered by the economists who model various economic growth aspects. Finally, polynomial math problems are used in healthcare to prevent the spread of bacteria among other things. While these may seem a bit complex, do not let them frighten you as we have good examples to help you achieve success.