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10

If all the coefficients are non-negative integers and if we know the max of all the coefficients, then it is easy to find the coefficients with just ${one\ point}$ of choice. Let's say the max is $M$. Then evaluate the polynomial at $M+1$. The output will be equal to the decimal number system representation of a number whose base $M+1$ number has digits as ...


10

Hint: First observe the equation is palindromic. Divide throughout with $x^2$ and rewite it as a quadratic in $\left(x+\dfrac{1}{x}\right)$.


8

Not knowing the substitution trick, you can anyway infer that if $x$ is a solution, then $1/x$ as well, so that the polynomial can be factored in two polynomials of the second degree, and these will be palindromic too: $$x^4 - 14x^3 + 50x^2 -14x + 1 =(x^2+Ax+1)(x^2+Bx+1).$$ Developing and identifying, $$A+B=-14,\\1+AB+1=50.$$ The solutions are ...


6

First, let's assume that Schanuel's conjecture is true and speak very loosely. :) Timothy Chow's 1999 article What is a Closed-Form Number? proves that the exponential and logarithm functions don't really help us to express algebraic numbers. Any algebraic number that can be expressed using those functions can also be expressed using only radicals. This is ...


6

With two points you can only uniquely determine $p(x)$ if it has degree one. In general if $p(x)$ has degree $n$, you will need the computer give $(n+1)$ outputs, each for a different input. For example if $p(x)$ has degree 2 and you input $0$ and $1$ and get outputs $a$ and $b$ respectively, there are an infinite number of parabolas which pass through ...


6

By the rational root test, the possible rational roots of the polynomial have the form $$ r=\frac{1}{k} $$ where $k$ is an integer (positive or negative) that divides $n$. Which of them are integer? When are they roots of the polynomial?


5

If you believe the question, that you are looking for one possible value of $n$, then you should be able to find one by setting $f(1)=0$ which clearly gives you a linear equation for $n$. Why doesn't $f(-1)=0$ work? You need something like @egreg's approach to show there are no other possibilities.


5

By the rational root theorem, if there is a rational root $p/q$ in lowest terms, then $p$ divides $a_{0}$ (i.e. $1$ is divisible by $p$, so what can $p$ be?). We can also say that $q$ divides $a_{3}=1$. What must $q$ be? Make sure you check whether the solution you get works!


5

Let $S(n)$ be the statement that " $P_n(x)$ has no real root for $n$ even, and has exactly one real roots for $n$ odd" You can check directly that $S(1)$ and $S(2)$ are true. Assume that $S(k)$ is true. Consider $k+1$-case: if $k$ is even, the induction hypothesis says that $P_k(x)$ has no real roots. As $P_k(0) = 1$, we have $P_k(x) >0$ for all ...


4

Hint $$x^4+4=(x^4\color{red}{+4x^2}+4) \color{red}{-4x^2}$$ and notice that $$a^2-b^2=(a-b)(a+b)$$


4

Using the fact that $a^3 + 1 = (a+1)(a^2 - a +1)$, we have $$(1-x)^3 + 1 = 0 \iff \Big((1-x)+1\Big)\Big((1-x)^2 - (1-x) + 1\Big) = (2 - x)(x^2 - x + 1)\\ \implies x = 2\, \text{ or }\, x^2 - x + 1 = 0$$ Use the quadratic formula on the remaining quadratic to solve for the remaining two roots.


3

Note: Let $f(x)=x^3-5x^2+11x+17$. $f(-1)=0\Rightarrow (x+1)|f(x)$. Divide, and get a quadratic factor. I'm sure you'll be able to find the other two roots then.


3

We can use the identities of Viète to avoid the hazards of polynomial division. The sum of the roots is $5$, so the sum of the missing roots is $6$. The product of the roots is $-17$, so the product of the missing roots is $17$. It follows that the missing roots are solutions of $x^2-6x+17=0$.


3

A rational function is zero when the numerator is zero, except when any such zero makes the denominator zero. $$f(x) = \frac{p(x)}{q(x)} = 0 \implies p(x) = 0 \text{ and } q(x) \neq 0$$ In this case, we need to solve $$2x^2 - 8 = 2(x^2 - 4) = 2(x-2)(x+2) = 0 \iff x = 2\text{ or } x= -2$$ Note that the denominator is not zero at either of those solutions. ...


3

It means that if $A(x) = 0$, then $B(x) \neq 0$, and if $B(x) = 0$, then $A(x) \neq 0$.


3

$x^4 + 4 = (x^2 + 2i)(x^2 - 2i)$ or without complex numbers : $x^4 + 4 = (x^2 - 2x + 2)(x^2 + 2x + 2)$ Maybe that helps you ?


3

I think the implication is true: Write $f=\prod\limits_{i=1}^nf_i^{r_i}$ for the irreducible factors $f_i$ of $f$ and $f+1=\prod\limits_{j=1}^m\tilde{f}_j^{s_j}$, respectively. Then, as in the comment above, one has that $\prod\limits_{i=1}^nf_i^{r_i-1}$ and $\prod\limits_{j=1}^m\tilde{f}_j^{s_j-1}$ divide the derivation $f'=(f+1)'$ and thus, as $f$ and ...


3

This follows from the Mason--Stothers theorem, which is the polynomial analogue of the $ABC$ conjecture. We need to be careful about stating the theorem in a way that works in characteristic $p$, as follows. Theorem (Mason, Stothers): If $k$ is a field and $a(x)$, $b(x)$, and $c(x)$ are nonzero polynomials in $k[x]$ such that (i) $a(x) + b(x) = c(x)$, ...


3

It seems to be recursively dividing using itself (the lead(r)/lead(d) line). "lead(r)" presumably means the leading term in the polynomial, so $x^2$ in $x^2+x+1$.


3

I can show the claim when the interval is the same for each integral, let's say it is $[-1,1]$, and if there is a restriction on the roots of $f(t)$. I thought about this for a while and figured I might as well post it. I'm slightly hopeful with some more thinking a full solution will come. First, suppose $f(t)=\prod (t-\alpha_i)$ with $|\alpha_i| > 3$. ...


3

I am assuming that by $\sin \theta_1 z^3$ our OP Jackie means $z^3 \sin \theta_1$ and so forth; with this understanding, we have the given equation $z^3 \sin \theta_1 + z^2 \sin \theta_2 + z \sin \theta_3 + \sin \theta_4 = 3; \tag{1}$ taking absolute values and using the triangle inequality (several times) yields $3 = \vert 3 \vert \le \vert z^3 \sin ...


3

Note that $$a^2-b^2=a^2-ab+ab-b^2=a(a-b)+b(a-b)=(a+b)(a-b)$$ Now put $a=(x-5)$ and $b=2$ $$(x-5)^2-4=(x-5)^2-2^2=((x-5)+2)((x-5)-2)=(x-5+2)(x-5-2)$$


3

A polynomial of degree $n$ has $n+1$ coefficients. If you know the degree but don't know anything about those coefficients, you'll need $n+1$ values of the polynomial to determine them. If you don't know the degree, no amount of values of the polynomial will suffice.


2

No. Unfortunately, determining if two multivariate polynomials have a common factor is quite a bit trickier than doing the same with single variable polynomials. Finding out whether $F$ and $G$ have a non-trivial common factor is generally hard to do by hand. However, there are techniques to handle this problem. The greatest common divisor (GCD) of $F$ and ...


2

Let $V$ be the subset of $P_3$ consisting of all polynomials of degree three and the zero polynomial. Note that $x^3+1,-x^3\in V$ while $$ x^3+1+(-x^3)=1\notin V $$ Hence $V$ is not a subspace.


2

To check whether a cubic polynomial is irreducible over a given field, it is sufficient to check whether $f$ has any roots in that field (why?). In $GF(7)$ this is easy: there are only $7$ possible roots, so you can simply evaluate $f(x)$ for all $x \in GF(7)$. Over $\mathbb{R}$, remember that every polynomial of odd degree has at least one root. You can ...


2

A general form for the cubic equation is, $$ax^3+bx^2+cx+d=0 \tag{1}$$ To find the roots of this equation we first try to get rid of the quadratic term $x^2$. The substitution $x=y-\dfrac{b}{3a}$ helps in achieving our goal. This results in, $$ay^3+\left(c-\dfrac{b^2}{3a}\right)y+\left(d+\dfrac{2b^3}{27a^2}-\dfrac{bc}{3a}\right)=0\tag{2}$$ which we ...


2

It's called a polynomial with $0$, $1$ coefficients, or a polynomial with coefficients from $\{0,1\}$. A polynomial with all coefficients equal to $1$ would be of the form $1+x+x^2+\cdots+x^{n-1}+x^n$.


2

Your notation for the field with $p$ elements is weird. Write $\mathbb{Z}/{p\mathbb{Z}}$ or $\mathbb{F}_p$ instead. By definition, the Frobenius automorphism is the polynomial function corresponding to the polynomial $X^p$ in $\mathbb{F}_p[X]$. Pay attention to distinguish between a polynomial and its associated polynomial function. Exercise: Any function ...


2

Write the coefficients of the elements as linear combination of the standard bases $\{1,x,x^2,x^3\}$ as rows in a matrix, then apply the elementary operations on the rows to get the matrix in row echelon form. Then you can read the rank of the matrix, which is equal to the number of independent elements in your set of polynomials. These elements correspond ...



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