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2

Mark's answer is nicer than this, but it is possible to continue along the lines you started: Since $f'(x)=(x^2+1)g'(x)+2x g(x)$, $2x g(x)$ has the same remainder $(x+1)$ when divided by $x^2+1$ as does $f'(x)$. That is, $$ 2xg(x)=h(x)(x^2+1)+x+1 $$ for some polynomial $h$. By setting $x=0$, we can see that $h(0)=-1$; thus $h(x)=xk(x)-1$ for some other ...


2

You can use the rule for a linear factor that $a$ is a double-root, and $(x-a)^2$ a factor, of $p(x)=0$ if and only if $p(a)=p'(a)=0$. Use $a=i$ and $a=-i$. $(x-i)^2(x+i)^2=(x^2+1)^2$


2

Using the Q.F., the roots of $$p(q) = q^2 + B q + C$$ are $$r_{\pm} = \frac{1}{2}\left(-B \pm \sqrt{B^2 - 4 C}\right)$$ (here $\sqrt{\cdot}$ is any branch). If $p$ has a double root, which the above shows happens iff $B^2 = 4C$, in which case the double root is at $-\frac{B}{2}$, then any reflection through a line containing $2$ maps each root to the other. ...


0

Those are all correct. Here's everything presented in a table: $$\begin{array}{lll} \textbf{element} & \textbf{reduced} & \textbf{min poly} \\ 0 & 0 & x \\ \alpha^0 & 1 & x+1 \\ \alpha^1 & \alpha & x^2+x+1 \\ \alpha^2 & \alpha^2 & x^3+x+1 \\ \alpha^3 & \alpha+1 & x^3+x^2+1 \\ \alpha^4 & \alpha^2+\alpha ...


0

set $(q-c)(q- (a+b))(q- (a-b))=0$. This should refer to the same polynomial. Compare the coefficient, we get $c+2a=3a$ $c(a+b)+c(a-b)+a^2-b^2=b^2$ $-c(a+b)(a-b)=c$ we get $a=b=c=0$


0

You can take the derivative and set to zero to get local maxima and minima (by calculus 1) which gives you $${n \choose k} (kx^{k-1}(1-x)^{n-k} - (n-k)x^k (1-x)^{n-k-1}) = 0$$ Clearly the maximum doesn't occur when $x = 0,1$ so you can divide by the common factors of $x$ and $(1-x)$ and you can also ignore the constant ${n \choose k}$ and then you get $$ ...


1

Let $\alpha$ be a root of $f(x)$ and let $m_\alpha(x)$ be the minimal polynomial of $\alpha$ over $K$. It's easy to conclude that $n=\deg\left(m_\alpha(x)\right)=[K(\alpha)\colon K]$. It holds that $K\preceq K(\alpha)\preceq F(\alpha)$ and $K\preceq F\preceq F(\alpha)$. Since all these extensions are finite it follows that $$\begin{align} ...


1

If $K$ is a finite field and $f \in K[X]$ is irreducible, then $K[X]/(f)$ is a splitting field of $f$. This follows from the fact that finite extensions of finite fields are always normal. If $\alpha$ is a root, the other roots are $\alpha^{p^n}$, $n \in \mathbb{N}$, where $p=\mathrm{char}(K)$. In particular, the degree is $\mathrm{deg}(f)$. If $n$ is a ...


0

Well, check again. For $y^2+(−1−x)y+x=0$ with $y$ as the quadratic variable. The quadratic formula with $b$ as $(-1-x)$, $a$ as $1$, and $c$ as $x$ tells that the roots are $$y_{1,2}=\displaystyle \frac{(1+x) \pm \sqrt{1+x^2+2x - 4x}}{2}=\frac{(1+x)\pm\sqrt{1+x^2-2x}}{2}=\frac{(1+x) \pm{1-x}}{2}$$ Hence the solutions are $$y_1=\frac{1+x+1-x}{2}=1 \ , \ ...


0

First fact: Polynomial equation of order $n$ has up to $n$ real solutions. So, expect $12$ solutions. If this appears in engineering as some model, and you know there has to be a unique inverse, you probably have very specific coefficients and very specific domain (if $x$ guaranteed to be in some range). Second fact: Polynomials with order above $4$ have ...


4

$$y^2-(x+1)y+x=0$$ First method - using the quadratic formula: $$y_{1,2}=\frac{x+1\pm\sqrt{(x+1)^2-4x}}{2}=\frac{x+1\pm\sqrt{(x-1)^2}}{2}=\frac{(x+1)\pm(x-1)}{2}$$ Thus the solutions are $y=x$ or $y=1$. Second method - taking common factors: $$y^2-yx-y+x=y(y-x)-1(y-x)=(y-1)(y-x)=0$$ Thus the solutions are $y=x$ or $y=1$.


2

You can complete the square to get $$\left( y-\frac{1+x}2\right)^2=\frac{(1-x)^2}4$$ from which the solutions are obvious.


5

Factor by grouping: $y^2 - yx - y + x = y(y-x)-(y-x)=(y-x)(y-1)=0$ So either $y-x=0$ or $y-1=0$. The result follows.


0

I don't really know much about about primitive polynomials But You can read about it here Following are primitive polynomials $$x^2+2x+2$$ $$x^2+x+2$$ and This Page Generates Primitive polynomials


3

Your approach is correct, but you should probably say that you let $\theta=\frac{\pi}{8}$ before you introduce $x$. Continuing after your third line, you could write something like this: Let $\theta = \frac{\pi}{8}$, and denote $x=\cos^2\left(\frac{\pi}{8}\right)$. This gives $$ \frac{\cos\pi-1}{32}=4x^4-8x^3+5x^2-x$$ so $4x^4-8x^3+5x^2-x=-\frac{1}{16}$.


4

Use the recursive relation involving Chebychev polynomials: $$ T_{n+1}(X) = 2XT_n(X) - T_{n-1}(X) \\ T_n(\cos \theta) = \cos n\theta $$ Proof: first prove existence of such polynomials. Then the identity $$ \cos ((n+1)\theta) + \cos ((n-1)\theta) = 2\cos \theta \cos n\theta $$


0

Let $$y=w-\frac{x}{w}$$ Then you get $$w^3-5x^3+1-\frac{x^3}{w^3}=0 \stackrel{\cdot w^3}{\iff} w^6+(-5x^3+1)w^3-x^3=0 \stackrel{\theta:=w^3}{\iff} \theta^2+(-5x^3+1)\theta-x^3=0 \iff \theta= \frac{5x^3-1\pm\sqrt{25x^6-6x^3+1}}{2} \iff y=\sqrt[3]{\theta}-\frac{x}{\sqrt[3]{\theta}}$$ Check your answers, the minus sign case on the quadratic doesn't satisfy ...


1

You have a depressed cubic equation $y^3+a y + b=0$ in which $a$ and $b$ are functions of $x$. So consider the test (Cardano method) $$4a^3+27b^2=675 x^6-162 x^3+27$$ has no real root and so it is positive. I am sure that you can take from here.


0

Edit: Originally the question asked was for the equation $y^2 + 3x - 5x^3 + 1 = 0$, and so this is a response to that question. How you can go about trying to understand this: Note that the given equation is a quadratic equation in $y$, whose coefficients are polynomials in $x$. That is, we can rewrite it as $$ y^2 + by + c = 0 $$ where $b = 3x$ and $c = ...


1

Answer to question before edit: It doesn't define $y$ as a function of $x$, because it doesn't pass the Vertical Line Test. Take for example $x=1$, which gives $y^2+3y-4=0 \Leftrightarrow y=1 \vee y=-4$.


1

Note that $\beta^4 + \gamma^4 + \delta^4 = S_4 - \alpha^4$, and similarly for the other quantities in parentheses. Substitute this in your desired expression, and expand it. You can then write the entire expression in terms of various $S$'s.


1

Hint: Consider any run of $n$ consecutive integers, say: $$ a + 1, a + 2, \ldots, a + n $$ Summing them together, we have: $$ \sum_{k=1}^n (a + k) = \sum_{k=1}^n a + \sum_{k=1}^n k = an + \frac{n(n + 1)}{2} = n\left( a + \frac{n + 1}{2}\right) $$ Now for what values of $n$ will the expression in brackets be an integer?


1

To compute the polynomial using a brute force approach, you can always use this : http://en.wikipedia.org/wiki/Lagrange_polynomial Your polynomial has degree $4$, so it will be a bit painful but you can expand it using a computer program if you're not up for it. Alternatively, the standard approach to compute this polynomial is using telescopic sums. ...


1

The question in your title is different than the question you ask at the end of your post. So I will answer them both. Firstly, it's not true that polynomials with rational coefficients have integer roots. For instance, $x^2 - 2$ has roots $\pm \sqrt 2$. However, it is true that polynomials of the form $p(x) = a_0 + a_1 {X \choose 1} + \ldots + a_d {X ...


1

Hint: compute $$ P(x+1) - P(x) $$ with $P(x) = ax^4 + bx^3 + cx^2 + dx$ and try to find $a,b,c,d$ so that $P(x+1) - P(x) = x^3$


0

Hint Note that $27^y=(3^y)^3$, $9^y= (3^y)^2$, if $x=3^y$ then: $$12 \cdot 27^y + 25 \cdot 9^y -4 \cdot 3^y -12=0\equiv 12x^3+25x^2 -4x -12=0$$


0

$$27^y = (3^3)^y = 3^{3 \cdot y} = 3^{y \cdot 3} = (3^y)^3$$


2

Use Euclid's algorithm in the usual way. The first step is $$(x^6+x^5+x^4+x^3+x^2+1)=(x^2)(x^4+x^3+x^2+1)+(x^3+1)$$ and the next will look like $$(x^4+x^3+x^2+1)=(\cdots)(x^3+1)+(\cdots)\ .$$


0

it is $x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-2x^2=(x^2+1-\sqrt{2}x)(x^2+1+\sqrt{2}x)$


1

It is the ideal generated by $\{x,y\}$. Explicitly, $(x,y)$ consists of those polynomials whose constant term is zero. Hint for the exercise: If $(x,y)=(g)$, then show that $g$ is a gcd of $x$ and $y$. But $x$ and $y$ are coprime.


1

Given a multivariate polynomial, a good place to start would be to factor out $x$ from all the terms with $x$, then factor $y$ out of all the terms that are left with $y$, and so on. Given a random multivariate polynomial of degree $n$ in $m$ variables, on average $O(n^m)$ terms will have an $x$, meaning just factoring out an $x$ will reduce the number of ...


0

Hint: the first condition gives $(x+1)^2$ as a factor, while the second gives $x-3$. So we must have $f(x)=a(x+1)^2(x-3)$ for some non-negative constant $a$. Use the remainder theorem to find $a$.


1

A function (such as in your example) does not necessarily have an inverse function. For a given value of $y$, there can be many values of $x$ for which $f(x) = y$. In the case of an irreducible degree $12$ polynomial, there will typically be $12$ different values of $x$ for which $f(x) = y$ for random choices of $y$. There are lots of choices of software ...


5

Unfortunately, most equations such as you give do not have a nice inverse. Consider a function like $y=x^2$. This has exactly one $y$-value for each $x$-value -- that's what makes it a function. What you are asking is about the inverse. You can visualize this by taking the graph of the original function, and swapping the roles of $y$ and $x$. The ...


0

One question that comes into mind is, what makes you thing that this can be done mathematically? Here is a different approach that may help you. Go into parametrics. So for example you have $y=x^{12}+2x^4 + 3$ First invert: $x=y^{12}+2y^4+3$ Now use parametrics: Set $y=t$ then $x=t^{12}+2t^4+3$ Maybe that helps for your purpose?


0

MathCad, its a great software which i also use.


0

Any quartic where terms of odd degree have zero coefficient,namely any quartic polynomial of form $f(x)=x^4+x^2+n$ where $n$ is any non-negative integer


0

So I think I found the answer using the conditition $gcd(r,p)=1$, which I didn't previously mention. It would be nice if someone could verify this answer. Since we are in $\mathbb{Z}[X]/(X^r-1,p)$, $X$ is a root of unity. But even better, since $gcd(r,p)=1$, $X^p$ is a root of $X$. $\implies \exists h : X^{ph} \equiv X$. From Edit3 we have ...


1

Hint: Substitute $q^5$ with $r$, so you get the equation $r^2-2r+2=0$


3

You have to solve it in two steps. Let $x=q^{5}$. Then $x²-2x+2=0$ Solve this, and then solve the first equation to find the solutions in $q$.


1

Step ii was done by long division: the polynomial $x^4 + 9x^3 + \ldots$ was divided by $x^2 + 10 x + 41$ to get $x^2 - x + 4$, with a remainder of $-160$.


2

It seems to be they simply divided with residue the two polynomials: $$x^4+9x^3+35x^2-x+4=(x^2+10x+41)(x^2-x+4)-160$$


1

Suppose you have $p(x)=(x-b)f(x)$ for some polynomial $f(x)$... Product of constant term of $f$ and $-b$ should be same as constant term of $p$. Thus, $b$ should divide constant term... So, it is enough to check for factors of $p(0)$..


1

But this require to check every b∈ℤ where ℤ is an infinite set. False. Only is required check the divisors ($+$ and $-$) of $p(0)$.


1

Suggestion. Instead of doing what you did for $$ f(\vartheta)=\frac{1}{1+\sin^2\vartheta}, $$ try in on each of the two terms of what $f(\vartheta)$ is equal to: $$ \frac{1}{1+\sin^2\vartheta}=\frac{i}{2}\left(\frac{1}{i+\sin\vartheta}+\frac{1}{i-\sin\vartheta}\right). $$


1

$\int^{\pi}_0 \frac{1}{1+ \sin^2 \theta}$ d$\theta =\int^{\pi}_0\frac{1}{1+\frac{1}{2}-\frac{\cos 2\theta}{2}}$ d$\theta = \int^{\pi}_0\frac{1}{\frac{3}{2}-\frac{\cos 2\theta}{2}}$ d$\theta = \int^{\pi}_0\frac{2}{3- \cos 2 \theta}$ d$\theta$ I believe that this can then be computed using Weierstrass substitution, and forming an improper integral. ...


0

All you know is this: a nonzero polynomial of degree $d$ cannot have more than $d$ roots in some (extension) field.


2

No. Take $p(X)=X(X+1)$, $F=\Bbb F_2$, $K=\Bbb F_4$. Indeed, let $\beta\in\Bbb F_4-\Bbb F_2$. We know that $\beta^2+\beta=1$, that is, $p(\beta)=1$. And $p(0)=p(1)=0$.


1

Hint: $$x^{n+1} - 1 = x(x^n - 1) + (x - 1)=x(x-1)(?) + (x - 1)=(x-1)(x(?) + 1)$$ Alternative: Actually We don't need to use Induction, Let $f(x)=x^n-1$ Now $f(1)=0$, That means $1$ is a root if $f(x)$ so $f(x)=(x-1)\times(?)$ ( By '$?$' I mean some another factors as $f(x)$ is polynomial of degree $n$ )


0

Yes, it is true that all unlinks have Alexander-Conway polynomial equal to zero. This comes from the more general fact that the Alexander polynomial of a splittable link is always 0. A link is splittable if its components can be seperated by a plane in $\mathbb{R}^3$, which is exactly what you would want it to mean. And the unlink obviously falls into ...



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