Hot answers tagged polynomials
10
Your previous question concerns generalized CRT. So you know that
$$\frac{{\bf R}[x]}{(x-1)(x^2+1)}\cong\frac{{\bf R}[x]}{(x-1)}\times \frac{{\bf R}[x]}{(x^2+1)}\cong {\bf R}\times{\bf C}$$
Find the idempotents in $\bf R$ and $\bf C$ to find the idempotents in ${\bf R}\times{\bf C}$, then pull them back through the isomorphisms implicit above.
Indeed ...
9
The following is correct, though not fully expanded:
$$(x^2 + 4x + 4)(x + 2) = (x+2)^2(x+2) = (x+2)^3$$
Consider writing this as $(x + 2)(x^2 + 4x + 4)$, and then distribute (multiply) each term in the first factor, with each term of the second factor.
$$
\begin{align} (\color{blue}{\bf x + 2})(x^2 + 4x + 4)
& = \color{blue}{\bf x}(x^2 + 4x + 4) + ...
7
Fun fact:
$$x^4+1=(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1).$$
This can be derived by setting $x^4+1$ equal to a product of two monic quadratics with unknown coefficients and then solving for said coefficients little by little. As a consequence,
$$x^8+1=(x^4+\sqrt{2}x^2+1)(x^4-\sqrt{2}x^2+1).$$
We can go further. For example, set
...
7
It is possible. For example, given a prime number $p$ and $a,b\in\mathbb Z$ with $a+b\le-2$, let
$$f(x)=x^3+pax^2+pbx+p\in \mathbb Z[x].$$
Then by Eisenstein's criterion, $f$ is irreducible in $\mathbb Q[x]$, i.e. $f$ has no rational root. However, since $f(0)=p>0$ and $f(1)=1+(a+b+1)p<0$, $f$ has three distinct real root located in $(-\infty,0)$, ...
6
For any polynomial $p(x)$, we have $\displaystyle\lim_{x\to\infty}\dfrac{p(x)}{e^x}=0$. This can be shown, for example, by using L'Hospital's Rule, and in other ways.
Your argument uses the derivative in a different way. The argument is somewhat informal, but it can be made formal. One could use induction. Let us prove by induction on degree that ...
6
Generally, if there's two roots whose sum is zero, then it means that two factors are $x-a$ and $x+a$, which means that $x^2-a^2$ must be a factor. So clearly
$$
(x^2-a^2)(x^2+bx+c)=x^4+2x^3-8x^2-18x-9=0
$$
Find the values of $a^2$, $b$, and $c$ that satisfy the left equality, and you'll have found factors that you can then solve for all roots. (This works ...
5
I doubt an easy proof of the irreducibility exists in general. If $n$ is a prime, then the polynomial is Artin-Schreier and handled easily.
Selmer gave a clever proof in the general case, working explicitly with the roots of the polynomial in $\mathbb{C}$. See E. S. Selmer, On the
irreducibility of certain trinomials, Math. Scand. 4 (1956), 287-302, ...
5
Why is $(x^{2} + 4x + 4)(x+2)$ wrong?
$(x^{2} + 4x + 4)(x+2) = x^{3} + 4x^{2} + 4x + 2x^{2} + 8x + 8 = x^{3} + 6x^{2} + 12x + 8$
And it is the right answer. If you have problem remembering the power formulas for binomials, just use pascal's triangle to calculate the coefficients:
\begin{align*}
(a + b)^{0}\to &1\\
(a + ...
4
Given: $(x+2)^3$ we can rewrite $(x+2)^3$ as: $(x+2)(x+2)(x+2)$
When expanding 3 terms we must first calculate $(x+2)(x+2)$ before we can multiply by the third $(x+2)$.
= $(x+2)(x+2)\implies x^2+2x+2x+4\implies x^2+4x+4$
= $ (x+2)(x^2+4x+4)\implies x(x^2+4x+4)$ + $2(x^2+4x+4)$
= $ x^3+4x^2+4x+2x^2+8x+8$
= $x^3+6x^2+12x+8$
4
Let $a$ be a root of $x^3+2x+1=0$ and $b=a^2$ be a root of the required equation
So, $a^3+2a+1=0\implies a\cdot b+2a+1=0\implies a=-\frac1{b+2}$
As $a$ be a root of $x^3+2x+1=0$, put this value of $a$ in $x^3+2x+1=0$
On simplification, I get $(b+2)^3-2(b+2)^2-1=0\iff b^3+4b^2+4b-1=0$
So, the required equation will be $y^3+4y^2+4y-1=0$
4
The condition for the cubic $x^3 + px + q$ to have 3 real roots is $4p^3+27q^2 \leq 0.$ This condition can be proved algebraically (search for Discriminants of a polynomial) or by calculus: A cubic has 3 real roots if and only if it has a local minimum and a local maximum and the product of those two values is $\leq 0.$
So as Potato mentioned in the ...
4
Hint: What values does the function $x^2$ acquire(positive/negarive)? What is the solution of the equation $x^2=0$?
Can you find the solution of the equation $x^2+y^2=0$?
Now, what can you say about the equation $(x-3)^2+(y-5)^2+(z-4)^2=0
$? Can you find the values of $x,y,z?$
4
The method you cite should work, because the polynomial is of degree $2$ and hence any factors will correspond to solutions of the equation $x^2+x+1=0$. But you have to do it right and work over the field $\mathbb F_{256}$, which means you must begin by constructing that field etc. Working over the integers modulo 256 (i.e. $\mathbb Z/256\mathbb Z$) will not ...
4
As "Myself" observed, you are apparently misconstruing $\mathbb F_{2^8}$ as being isomorphic to $\mathbb Z/2^8$, which it is not. For example, the latter has many zero-divisors, and is far from being a field.
Also, in any case, for human-executable work checking 256 cases is usually a terrible method. Instead, look for some structure or meaning. Here, the ...
4
Steps:
Prove that constant functions are continuous
Prove that the identity function $f(x)=x$ is continuous
Prove that if $f(x)$ and $g(x)$ are continuous, then so are $f+g$ and $f\cdot g$.
This suffices to prove that all polynomials are continuous.
4
Using middle term factor, $$(x^2+5)^2-(9+6)(x^2+5)+6\cdot9=0$$
$$\implies (x^2+5)(x^2+5-9)-6(x^2+5-9)=0$$
$$\implies (x^2-4)(x^2-1)=0$$
$\implies x^2-1=0$ or $x^2-4=0$
Alternatively, using quadratic equation formula for $x^2+5=\frac{15\pm\sqrt{15^2-4\cdot1\cdot 54}}{2\cdot1}=\frac{15\pm 3}2=9$ or $6$
4
Let $a = (x^2 + 5)$. Then $$(a-9)(a-6)=0$$
$$\implies a = 9\; or \;a = 6$$
$$\implies x^2+5 = 9 \; or \;6$$
$$\implies x = \pm\sqrt4\; or \pm\sqrt1$$
$$\implies x = \pm2\; or \pm1$$
This kind of equation is called a biquadratic equation. Cheers!
3
First: looking for roots to deduce a polynomial over some field is irreducilbe works as long as the polynomials degree $\,\le 3\,$ .
Second: the polynomial $\,x^2+x+1\,$ is defined over $\,\Bbb F_2\,$ and it's irreducible over this field. Since $\,2\mid 8\;$, we have that $\,\Bbb F_{2^2}\le\Bbb F_{2^8}\,$ , and this means that all the roots of $\,x^2+x+1\,$ ...
3
No error on your part:
If you've copied the problem correctly, then your solution is correct: $k = -3$.
I was very careful in calculating, as I'm sure you were, in double checking, so if $(x - 2)$ is a factor for your given polynomial, then $k$ must be $-3$.
Typo/misprint I suspect, in your text: a typo in the solution, or a misprint of the desired ...
3
Hints:
The constant coefficient (constant term) is $-9$.
$$\pm 1, \pm 3\pm 9 \;\text{divide}\; -9 $$
If there are rational roots to the polynomial, they will be among these values. $\large (\star)$
Two of the roots $x_i, x_j$ be must be factors of $-9$: and they must be such that $x_i = -x_j$. Lets call this pair $a, \pm a$. That means $(x−a)$ and ...
3
This is a standard example of an equation
that does not have an explicit formula for its root.
It can only be solved numerically.
The important thing is to get a good first approximation to the root.
For example, if you want to solve
$\frac{(1+x)^n-1}{x} = a$,
and you think that there might be a root close to $0$,
you can approximate $(1+x)^n$
by ...
3
The first expression evaluates to $e^{2u}-e^u-e^l+1$, so if your question is whether there is some algebraic manipulation that brings this into the form of an exponential minus $1$ for general $l,u$ then the answer is no.
In fact the range of the function $x\mapsto e^x-1$ in $\Bbb R$ is $(-1,\infty)$, and the product of two values in that range may be ...
3
The answer depends on $R$.
Example $1$: $R=\mathbb{C}$ and you know that any polynomial, including
the one in question, splits into linear factors
Example $2$: $R=\mathbb{Q}$ denote $$f(x)=x^{8}+1$$ then $$f(x+1)=(x+1)^{8}+1=x^{8}+8x^{7}+28x^{6}+56x^{5}+70x^{4}+56x^{3}+28x^{2}+8x+2$$
is irreducible by Eisenstein with $p=2$ hence $f(x)$ is irreducible.
2
Over $\mathbb{C}$, this polynomial splits into linear factors of 8 of the 16 solutions to $x^{16} - 1 = 0$. The irreducible factors of this equation are the cyclotomic polynomials.
http://en.wikipedia.org/wiki/Cyclotomic_polynomial
So $x^8 + 1$ is indeed irreducible over $\mathbb{Q}$, as it is a cyclotomic polynomial.
Over $\mathbb{R}$, this polynomial ...
2
Considering $\sigma(A)$ denotes the set of characteristic(eigen) values of A
As $\lambda$ is an eigen value, $\exists 0\ne v\in V$ such that $Av=\lambda v$
Note that $A^kv=\lambda^k v,\forall k\in N$(easy to prove using Induction)
Let $p(x)=\sum_{i=0}^{n}a_ix^i,a_i\in R$
Then we have ,
$p(A)=\sum_{i=0}^{n}a_iA^i$
So we have ,
...
2
If $\,f\,$ is a polynomial in $\,n\,$ indeterminates with coefficients in $\,R\,$ , we want $\,\phi f\,$ to be an element in $\,S\,$ , and that's why there's written $\,\phi f(s_1,...,s_n)\;,\;\;s_i\in S\,$ . For this to happen, we need the coefficients of the polynomial $\,f\,$ in $\,R[x_1,...,x_n]\,$ to be mapped to elements in $\,S\,$ , and this is ...
2
Make a change in variables: $y = x + 1$. Then the equation becomes:
$$\dfrac{y^{36} - 1}{y-1} = 20142.9/420$$
The numerator $y^{36}-1$ is equal to $(y-1)(y^{35}+y^{34}+\cdots+1)$ (use the formula for the sum of a geometric series). Therefore dividing by $y-1$ gives:
$$y^{35}+y^{34}+\cdots+1 = 20142.9/420$$
This is a polynomial of degree 35 in $y$. It ...
2
Edit: There is a mistake in this proof. I think it can be fixed with an argument from the degrees of $q$ and $q'$, so I'll work on that.
Suppose $p(x)$ is reducible. Then
$$
p(x) = (x^i + q(x))(x^j+q'(x))
$$
with $i+j=n$. But then $$x^iq'(x) + x^jq(x) + q(x)q'(x)=-x - 1.$$
forces
$$
x^iq'(x) + x^jq(x) = -x
$$
$$
q(x)q'(x) = -1.
$$
Hence, $q(x), q'(x)$ are ...
2
Solved the equation using Tartaglia's method
$$ x^3 + x^2 + 1$$
Make then substitution $ x = t + h $
$$ t^3 + 3t^2h+3th^2 + h^3 + t^2 +2th+h^2 + 1 = 0$$
$$ t^3 + t^2(3h+1)+t(3h^2+2h) + (h^3 + h^2 + 1) = 0$$
In order to eliminate the second degree term we add a condition $ h = -\frac{1}{3}$
$$ t^3 -\frac{1}{3}t + \frac{29}{27} $$
We make $t = u+v$
$$ u^3 ...
2
It means the roots are of the form $a, ar, ar^2$.
Here it is not too difficult to see that $2, 4, 8$ are ok, just look at the constant term, which is $- (a \cdot ar \cdot ar^2) = - (a r)^3$, and check that $2, 4, 8$ fit with the other coefficients
$$
14 = 2 + 4 + 8,
\qquad
56 = 2 \cdot 4 + 2 \cdot 8 + 4 \cdot 8.
$$
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