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1

You have to make the difference between (formal) polynomials and polynomial functions. A rigourous definition of a polynomial over a commutative ring $A$ is that, as a set, it is $$A^{(\mathbf N)}=\bigl\{(a_0,a_1,\dots, a_k,\dots)\mid a_k\in A \text{ and all $a_k$ but a finite number are 0}\bigr\}$$ In other words, a polynomial is identified with the ...


0

When you see the polynomial as an object, then $x$ is a "placeholder". But you can also see the polynomial as a function, where you think of $x$ as a variable. The two notions agree (in the sense that if two polynomials are equal as functions then they are equal as polynomials) when the coefficient ring is $\mathbb Z$, $\mathbb Q$, $\mathbb R$, or $\mathbb ...


0

By setting $y=x-1$, we just have to find the remainder of $(y+2)^n$ when divided by $y^3$. By the binomial theorem: $$(y+2)^n \equiv 2^n + n2^{n-1} y + n(n-1)2^{n-3} y^2\pmod{y^3}.$$


2

You can write the equality (E): $$(x+1)^n=P(x)(x-1)^3+a(x-1)^2+b(x-1)+c$$ where $P(x)$ is the quotient and $a(x-1)^2+b(x-1)+c$ the reminder of the euclidean division of $(x+1)^n$ by $(x-1)^3$. And your problem is to find $a,b,c$. (E) can be seen as an equality between functions. So can you differentiate both side to get further equalities. Make $x=1$ in ...


0

By the remainder theorem, the remainder when $p(x)$ is divided by $g(x)$, is $p(a)$ for $g(a)=0$. So in the above problem, $a=1$, therefore your answer should be $2^n$


5

$$ (x+1)^n=((x-1)+2)^n=2^n+n 2^{n-1}(x-1)+\binom{n}{2}2^{n-2}(x-1)^2+(x-1)^3 \cdot F(x) $$ So the remainder is $$2^n+n 2^{n-1}(x-1)+\binom{n}{2}2^{n-2}(x-1)^2.$$


3

Let $A$ and $B$ be domains with $A\subseteq B$. We say that the ring extension $B/A$ is good if the following implication holds: $$\text{For every}\ f\in A\setminus\{0\}\ \text{and}\ h\in B,\ \text{if}\ fh\in A\ \text{then}\ h\in A\,.$$ We claim that if $B/A$ is good then $B[x]/A[x]$ is good as well. In fact, let $f\in A[x]\setminus\{0\}$ and $h\in ...


0

The easiest way I've found to get the equation for a bezier curve of order / degree $N$ is to use the binomial theorem and pascal's triangle. Basically, the generic formula for Bezier curves looks like $(As+Bt)^N$ where $N$ is the degree of the curve which has $N+1$ control points. You could expand this by hand. If you wanted to create a quadratic curve ...


1

Write $g=fq+r$, with $q,r \in K[x]$ and $r=0$ or $\deg(r)<\deg(f)$. Write $g=fh$, with $h \in L[x]$. Then $r = f(h-q)$. If $r\ne0$, then $h-q\ne0$, and so $\deg(f(h-q))\ge \deg(f) > \deg(r)$. Thus, $r=0$. The only detail is that the degree of a polynomial in $K[x]$ remains the same when it is considered as a polynomial in $L[x]$.


2

Let $f\mid g$ in $L[x]$. Polynomial long division yields a polynomial $q\in L[x]$ with $f = gq$. A closer look at the polynomial long division algorithm shows that the coefficients of $q$ are computed by repeatedly applying field operations to the coefficients of $f$ and $g$. So if the coefficients of $f$ and $g$ are in $K$, then the computed coefficients of ...


0

Here's a more elementary proof. It's more cumbersome than other answers here, but it gets the job done... Assume there are three distinct roots of your quadratic $r_1, r_2, r_3$. Then we have \begin{align*} ar_1^2 + br_1 + c &= 0\\ ar_2^2 + br_2 + c &= 0\\ ar_3^2 + br_3 + c &= 0 \end{align*} Subtracting the second equation from the first, we get ...


1

This is not true in general. The polynomial $x^2+1$ has more than two zeroes over the quaternions $\mathbb{H}$, but is not identical zero. I suppose you assume implicitly that the domain is a field ? Over a field every nonzero polynomial of degree $n$ has at most $n$ zeroes, see here. The proof uses the Vandermonde matrix. Hence if a quadratic polynomial has ...


6

Let the three roots be $x_1,x_2,x_3$. Method $1$: Let $f(x) = ax^2+bx+c$. Since $x_1$ and $x_2$ are roots, this means $f(x) = (x-x_1)(x-x_2)g(x)$. Since $f(x)$ has degree $2$, this forces $g(x)$ to be a constant say $k$. Further, we have $f(x_3) = 0$. This means $k(x_3-x_1)(x_3-x_2) = 0$. Since $x_3 \neq x_1$ and $x_3 \neq x_2$, this forces $k$ to be zero. ...


0

This has already been answered previously on SE. The proof uses Chebyshev polynomials. Proof of a lower bound of the norm of an arbitrary monic polynomial


2

Hint:since $$a+b+c=1,ab+bc+ac=\dfrac{1}{3}$$ so $$(a+b+c)^2-3(ab+bc+ac)=0\Longrightarrow (a-b)^2+(b-c)^2+(c-a)^2=0\Longrightarrow a=b=c$$


0

Upto an integer scalar multiple, $P_n(x)$ is the truncation of $e^x$. It was proved by Schur that this polynomial is irreducible (cannot be factorized as product of lower degree polynomials with rational coefficients). T N Shorey has worked in generalizing this result. It will have results relevant to your quest.


0

After Jack's comment, it seems that $$P_n(x)=e^x \,\Gamma (n+1,x)$$ match the expression so $$P_{n+1}(x)=\frac{\Gamma (n+2,x)}{\Gamma (n+1,x)}\,P_n(x)$$ Now, the name ?


0

$$p(x)=ax^2+bx+c$$ given that $x=1+\sqrt 3$ so there must me an $x=1- \sqrt 3$ as in a quadratic polynomial when you get the roots the irrational part is from the discriminant as $a,b,c$ are rational. $$p(x)=(x-(1+\sqrt 3))(x-(1- \sqrt 3))$$ expanding this we get, $$p(x)=x^2-2x-2$$


4

Hint: If one of the roots is $1 + \sqrt{3}$ and $a, b, c$ are rational then what must the other root be? Solution 1: The other root must then be $1 - \sqrt{3}$. You have two options now: You can make a system of three linear equations in three variables and simply solve. Or, you can simply expand $\alpha[x - (1 - \sqrt{3})][x - (1 + \sqrt{3})] = ...


-1

Let $x = 1+\sqrt{3} \Rightarrow (x-1)^2 = 3 \Rightarrow x^2-2x-2 = 0 \Rightarrow a = k, b = -2k,c = -2k$, and $p(2) = -2 \Rightarrow 4a+2b+c=-2\Rightarrow 4k - 4k - 2k = -2 \Rightarrow k = 1 \Rightarrow (a,b,c) = (1,-2,-2)$.


1

Maple does the job by factor(expand((x^2+y^2+z^2)*(x+y+z)*(x+y-z)*(-x+y+z)*(x-y+z)-8*x^2*y^2*z^2)); $$ - \left( {x}^{2}-{y}^{2}-{z}^{2} \right) \left( {x}^{2}+{y}^{2}-{z}^{ 2} \right) \left( {x}^{2}-{y}^{2}+{z}^{2} \right) .$$


4

If $g$ is a polynomial of degree $m$ with leading coefficient $a$, i.e. $g(x) = a x^m + \ldots$ where $\ldots$ consists of terms of lower order, and $m \ge 1$ then $f(g(x)) = 2013 a x^m + \ldots$ while $g(f(x)) = 2013^m a x^m + \ldots$, so $f(g(x)) - g(f(x)) = (2013 - 2013^m) a x^m + \ldots$. Thus $f(g(x)) - g(f(x))$ can't be $0$ unless $m \le 1$. If we try ...


1

Let $\,a=x\!-\!1, b=x\!-\!2.\,$ Squaring a Bezout Identity (BI) for $\,\color{#c00}{a,b}\,$ yields a BI for $\,\color{#0a0}{a^2,b^2}$ $$1 = \color{#c00}{a-b}\overset{\rm square}\Rightarrow 1 = a^2\!+b^2\!-2ab(\color{#c00}{a\!-\!b})\ =\, (1\!-\!2b)\,\color{#0a0}{a^2}\!+(1\!+\!2a)\,\color{#0a0}{b^2}\qquad $$ Finally, from the BI for ...


0

You could say that $P(1)=2$, $P'(1)=2$, $P(2)=6$, $P'(2)=3$. This is four equations in the coefficients of $P(x)$. Let $P(x)=Ax^3+Bx^2+Cx+D$, and you have four equations in four coefficients. For example, $P(1)=A+B+C+D=2$.


0

Not sure, whether this is necessary (meaning, that there is probably different more general approach - as shown in the other answer. But if we want to use the identity of the degrees we need to split it), but let's divide it into two cases. First when the degree of $f, g$ is $0$ and then else: In the first case $\operatorname{deg}(f(x)) = 0$ and ...


4

You can see here more generally: Theorem. If the coefficient ring $R$ is an integral domain, then so is also its polynomial ring $R[X]$. Let $f(X)$ and $g(X)$ be two non-zero polynomials in $R[X]$ and let $a_f$ and $b_g$ be their leading coefficients, respectively.  Thus  $a_f≠0,  b_g≠0$,  and because $R$ has no zero divisors,  $a_fb_g≠0$.  But the ...


0

I think here for factoring the polynomial $x^6+5x^3+8$ we must go ahead by the same method of Eric Tressler. But, when we get the roots, it is seen that two complex roots are conjugate and so form the factor $x^2-x+2$. Hence, the factorization of Veritas is obtained. I think with this way Veritas get the factorization.


1

Let $E: y^2=f(x)$ and $\psi_3(x) = 2f(x)f''(x)-(f'(x))^2$, where $f(x)$ is a monic cubic polynomial with three distinct roots (because $E$ is non-singular!), as above. This can be shown using the following hint: A quartic polynomial $p(x)=a_4x^4+\cdots+a_0$, with $a_4>0$, has exactly two real roots if $p(x)$ takes negative values at all the zeros of ...


2

Even easier, $1-1^{2^{n-1}}=0$, so $1-x^{2^{n-1}}=(x-1)P(x)$ for some polynomial $P$. We don't need a constructive solution ;)


0

We have $$\forall m\in \mathbb{N}, 1-x^m=(1-x)(1+x+\cdots+x^{m-1})$$ Just take $m=2^{n-1}$ and you're done


4

Hint: for any $k\in\Bbb N$ we have $1-x^k=(1-x)(1+x+x^2+x^3+...+x^{k-1})$.


2

Answer summary: The generating function $r(u) = 1 + 3 u + 21 u^2 + 183 u^3 + 1773 u^4 \cdots$ is the unique solution to $$ 4 - 3 r(u)^2 - r(u)^3 + 27 r(u)^3 u=0 \qquad (\ast) $$ with $r(0)=1$. We can use this formula to generate many terms quickly, and to determine the asymptotic behavior of the sequence. To a first approximation, the sequence grows like ...


0

I think I got a solution so I will share it so you can give me your opinion: We know that this quartic polynomial will have at least two real roots , suppose we have 4 roots $\{ \alpha_1,\beta, \gamma,\alpha_2 \}$ conveniently orderded $\alpha_1< \beta <\gamma< \alpha_2$. Note that $f''(x)=6x+2a$ and $\psi_3'=12f(x)$ Since $\psi_3(\alpha_1)=0 ...


0

The function is $f(x)=\begin{cases} 2,\ x\neq 2 \\ \text{undefined}, \ x= 2 \end{cases}$ The function is discontinous. The empty circle shows, that the function is undifined at x=2. The graph is


3

When you start with a function like $f(x) = \dfrac{2x-4}{x-2}$, you should take note of any restrictions to the domain; in this case, $x \neq 2$. Now, whatever manipulation you do from this point on has an implicit condition that $x \neq 2$: $$ \begin{align} f(x) &= \dfrac{2x-4}{x-2}, \quad x \neq 2\\ &= \dfrac{2(x-2)}{x-2}, \quad x \neq 2\\ &= ...


0

A "generalized polynomial" with rational exponents can be turned to an ordinary polynomial by the change of variable $x=t^g$, where $g$ is the least common multiple of the denominators of all exponents. Example: $$4x^2-2\sqrt[3]x+3x^{4/7}=0$$ becomes $$4t^{42}-2t^7+3t^{12}=0$$ which has $42$ (possibly complex) roots in $t$. Things are less easy for ...


-1

when $x=2$ we get $$\frac{4-4}{2-2}$$ this is undefined and we have $f(x)=2$ if $x\ne 2$


1

Actually, the way stated your $f(x)$ is undefined at $x=2$ because of the zero in the denominator. There is a continuous extension of $f(x)$, say $c(x)$, which would indeed satisfy $c(2)=2$.


3

You mean something like $x^{\sqrt 7}+4x^2+x^{1/2}$? That's not usually called a polynomial at all -- it is implicit in the word "polynomial" that all of the exponents must be constant nonnegative integers. A sum of terms of this form will generally only make sense for positive $x$. The number of positive real roots can still be bounded using Descartes' rule ...


0

The only obstruction to the existence of a holomorphic square root is whether $\text{Arg}h(z)$ is well-defined up to an integer multiple of $2\pi$ given $\text{Arg}p(z)$. We shall show that for $|z|>R$, $\text{Arg}p(z)$ can be defined up to an even multiple of $2\pi$. Then it follows that $\text{Arg}h(z)$ is defined up to an integer multiple of $2\pi$, ...


0

Not all polynomials reducible over a field $F$ can be written as a product of linear factors. For example, the factorization of $x^3 - 1$ over $\Bbb Q$ into irreducible polynomials is $$x^3 - 1 = (x - 1)(x^2 + x + 1),$$ in particular, into a product of a linear factor and a quadratic one. Your proof doesn't require a factorization into linear factors, ...


3

By definition an element $f$ of a ring $R$ is irreducible if whenever $f=uv$ for $u,v\in R$, either $u$ or $v$ is a unit. By definition if $f$ is nonprimitive in $\Bbb{Z}[x]$, $f=ng(x)$ where $n$ is the gcd of it's coefficients, which is not $\pm 1$, and hence not a unit, and $g(x)$ is a primitive polynomial in $\Bbb{Z}[x]$ of positive degree and hence not ...


1

If the modulus is small, you should be able to do this by trial and error. For example, $x=4^{-1}$ modulo $7$ means $$4x\equiv1\pmod7\ ,$$ and you should easily be able to see that this is the same as $$4x\equiv8\pmod7\ ,$$ giving $x=2$. Another example: find $5^{-1}$ modulo $11$. We have $$5x\equiv1\equiv45\pmod{11}\ ,$$ so $x=9$. If it's too hard to do ...


12

$$(x^2-4)(x^2-2x)=2$$ $$\Rightarrow x^4-2x^3-4x^2+8x-2=0$$ $$\Rightarrow (x^2-x-1)^2-3(x-1)^2=0$$ $$\Rightarrow (x^2-x-1+\sqrt 3\ (x-1))(x^2-x-1-\sqrt 3\ (x-1))=0$$ $$\Rightarrow x^2+(\sqrt 3-1)x-1-\sqrt 3=0\ \ \text{or}\ \ x^2-(\sqrt 3+1)x-1+\sqrt 3=0$$ $$\Rightarrow x=\frac{-\sqrt 3+1\pm\sqrt{8+2\sqrt 3}}{2},\frac{\sqrt 3+1\pm\sqrt{8-2\sqrt 3}}{2}.$$


0

You want the second paragraph in that section, he explains it there. He is trying to use the points $-\pi/3, -\pi/6, \pi/6,$ and $\pi/3$ to generate $p_3(x)$, a cubic approximating the sine. Find a polynomial-part that takes the right value at $-\pi/3$ and is $0$ at $-\pi/6, \pi/6$ and $\pi/3$. Do the same for the other three points. Add these together. ...


2

When we do Polynomial Long Division, we are actually using Euclidean division algorithm. We are finding $q$ and $r$ such that $$a=bq+r$$ and $$\deg(r)<\deg(b)$$By definition, it is stated that we are only interested in solutions where the divisor has a greater degree, even though other solutions do exist. Therefore, if $\deg(b)=1$, we have $\deg(r)=0$.


2

When you divide by a number, the remainder should always be less than that number - otherwise, you could "put in" one more: $27 : 5 = 5\:\mathrm{rem}\:2$, not $4\:\mathrm{rem}\:7$. Similarly, when you divide by a polynomial, the remainder should always be less (in degree) than your divisor polynom: $\frac{x^2+3x+1}{x+1}=x+2-\frac{1}{x+1}$, not ...


2

If $f\in R[X]$ and $r\in R$ divides $f$, then $r$ divides every coefficient of $f$, hence the content of $f$, which is a greatest common divisor of the coefficients of $f$.


0

The content of a polynomial is the ideal generated by its coefficients. If $f= g h$ where $g$ is an element of the ring then this means that each coefficient of $f$ is a multiple of $g$, and the claim follows.


1

Whole point is that if $\lambda_j$ is a root then $\bar{\lambda}_j$ is also a root. By the choice of $\lambda_j$, $\lambda_k$ is not real. And the quadratic he considers is something which divides the original polynomial $f$. Ie the quadratic divides $f$ and has degree strictly less than $f$ so $f$ is reducible by definition.



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