For questions about finding factors of e.g. integers or polynomials

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2
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0answers
37 views

Factoring $x^5+B x^4+C x^3+D x^2+E x+F=(x^2+a x+b)(x^3+p x+q)$ over $\mathbb{Q}$

For a quntic polynomial to be reducible to the following form over $\mathbb{Q}$: $$x^5+B x^4+C x^3+D x^2+E x+F=(x^2+a x+b)(x^3+p x+q)$$ We need to match the coefficients ($a=B$ obviously, so we ...
3
votes
4answers
86 views

Factorize a third degree polynomial

It's my first time posting here so I'm not used to describing my problem in mathematics. I'm currently trying to solve a problem which asks if a 3x3 matrix is diagonalizable, I know the method but ...
1
vote
4answers
66 views

Why is $-32^{\frac{1}{5}} = 2$

When you factorize $-32$, you get: $-32 = (-16) \cdot 2$ $-16 = (-8) \cdot 2$ $-8 = (-4) \cdot 2$ $-4 = (-2) \cdot 2$ $-32^{\frac{1}{5}} = -2$ The reason I am asking is because you get $-4 = -2 \...
0
votes
4answers
112 views

Factorizing Faster

As one of the steps in a calculation, I had to factorize the following quadratic polynomial: $$12x^2 - 11x - 15$$ Using guess and check, and after roughly 10 minutes, I came to the following ...
0
votes
1answer
25 views

Factoring GCF from squared quantities

hope you're all well. Quick question: Am I allowed to factor this the way I did here? Thanks for the help, as always! :)
1
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2answers
55 views

$f(x+a)$ irreducibility means $f(x)$ irreducibility

Let $a~\in~\mathbb{Z}$ and let $f(x)~\in~\mathbb{Z}\left[x\right]$. Suppose that $f(x+a)$ is irreducible over $\mathbb{Z}$. Prove that $f(x)$ is irreducible over $\mathbb{Z}$. My idea is: $f(x)=u(x)*...
0
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0answers
23 views

Factoring polynomials when roots are external to the ring

I shall avoid maths script since I'm typing on a mobile, anyway I think I can do without. I have a question about factoring polynomials over a ring. Let's call R the ring in question. It is clear to ...
0
votes
1answer
9 views

Why is the running time of the trial division $O(f \cdot (log N)^2)$?

I saw this being cited in a few paper,but none of them seems to explain why this is the case. Maybe because it is quite trivial, but I am not sure why exactly... Here $f$ is the size of the factor. I ...
2
votes
0answers
29 views

On equivalence of RSA and factoring [duplicate]

Suppose we are given a number "$A$" which is multiple of $\phi(n)$. One can assume factorization to be hard. So you cannot find exact value of $\phi(n)$ from $A$. Clearly using this we can crack ...
2
votes
2answers
36 views

How to divide the following polynomial and factor it?

The question is $$ (2x^3+3x^2-39x-20) / (x-4) $$ I divided the following and got this as the answer $$ 2x^2+9x+3-8/(x-4))$$ I thought that this was the answer, but when i looked at the answer sheet ...
1
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1answer
63 views
-2
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1answer
28 views

What is the algorithm to factor something like $2+\frac{1}{x}+x?$ [duplicate]

I came across this in homework but I'm interested in the general example, say $ax+bx^{-1}+c.$
1
vote
2answers
49 views

Factoring $p(x) = x^n -1$ for any natural number $n$

Can I say that from inspection, $(x-1)$ is a factor which implies that $p(1) = 0$. I then used long division which then gave this: $p(x) = (x-1)(\sum \limits _{i=1} ^{n} x^{n-i})$ How would you have ...
1
vote
0answers
14 views

Is it easy to factor if we know $k\phi(PQ)$?

Suppose we know $k\phi(N)=k\phi(PQ)=k(P-1)(Q-1)$ where in $N=PQ$ we have $P,Q$ being similar sized primes and $k\in\Bbb Z$ is unknown can we factor $N$ in polynomial time?
1
vote
1answer
17 views

Prove that $a(x)$ divides $(v(x) - t(x))$

"Let $a(x), b(x) \in \mathbb{R}[x]$, not both the zero polynomial and suppose that gcd[$a(x), b(x)$] = 1. Let $u(x), v(x) \in \mathbb{R}[x]$ be such that $a(x)u(x) + b(x)v(x) = 1$ Let also $s(x)t(x) ...
0
votes
2answers
16 views

Prove that $q(x)$ does not divide $p_k(x)$

Let $n \in \mathbb{N}$ and let $p_1(x), p_2(x), ... *p_n(x)$ be $n$ irreducible polynomials over $\mathbb{R}$. Define the polynomial $p(x) = p_1(x) * p_2(x) *... *p_n(x) + 1 $ where 1 is the constant ...
4
votes
10answers
156 views

Factor $6x^2​ −7x−5=0$

I'm trying to factor $$6x^2​ −7x−5=0$$ but I have no clue about how to do it. I would be able to factor this: $$x^2-14x+40=0$$ $$a+b=-14$$ $$ab=40$$ But $6x^2​ −7x−5=0$ looks like it's not ...
1
vote
2answers
66 views

Factoring the sequence ${10}^{2n}+10^{n}+1$

While I am waiting for the basketball NBA game between Cleveland Cavaliers and Golden State Warriors to begin I sort of played with the sequence $a_n={10}^{2n}+10^{n}+1$ in a way that I looked for the ...
3
votes
3answers
105 views

Need help solving $x^4-3x^3-11x^2+3x+10=0$

Solve $x^4-3x^3-11x^2+3x+10=0$ I have tried to solve this equation using 'general formula from roots' from https://en.wikipedia.org/wiki/Quartic_function. $$ax^4+bx^3+cx^2+dx+e=0$$ $$x_{1,2}=-\frac ...
0
votes
1answer
21 views

Solving a characteristic Polynomial of the Hilbert Matrix

I need to find the eigenvalues of the following characteristic polynomial but I can't seem to successfully find the roots of the equation: $P[λ]$ = $λ^5$ - $563/315λ^4$ + $0.3476λ^3$ - $0.0038λ^2$ ...
2
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0answers
30 views

Books about multivariate polynomials

I'm looking for a book on multivariate polynomials, preferably a monograph (could also be a chapter inside another book). I'm interested in what can be said about roots, factoring, irreducibility, ...
92
votes
12answers
5k views

Why does factoring eliminate a hole in the limit?

$$\lim _{x\rightarrow 5}\frac{x^2-25}{x-5} = \lim_{x\rightarrow 5} (x+5)$$ I understand that to evaluate a limit that has a zero ("hole") in the denominator we have to factor and cancel terms, and ...
4
votes
1answer
39 views

How many positive two-digit integers have exactly 8 positive factors?

I solved this problem by listing all two-digit integers and going through each one. Is there an easy way to solve the problem? How many positive two-digit integers have exactly 8 positive factors?
0
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3answers
50 views
4
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3answers
4k views

Factoring the quintic polynomial $x^5+4x^3+x^2+4=0$

I am trying to factor $$x^5+4x^3+x^2+4=0$$ I've used Ruffini's rule to get $$(x+1)(x^4-x^3+5x^2-4x+4)=0$$ But I don't know what to do next. The solution is $(x+1) (x^2+4) (x^2-x+1) = 0$. I've ...
11
votes
3answers
1k views

A theorem about prime divisors of generalized Fermat numbers?

A theorem of Édouard Lucas related to the Fermat numbers states that : Any prime divisor $p$ of $F_n=2^{2^n}+1$ is of the form $p=k\cdot 2^{n+2}+1$ whenever $n$ is greater than one. Does anyone ...
2
votes
1answer
72 views

Finding 8 co-primes $\le 2^n$

We can find 8 co-prime integers $\le 2^n$ for sufficiently large $n$. I'm looking for asymptotic bounds for the minimum distance away from $2^n$ we have to go before finding 8 co-primes. In other ...
0
votes
1answer
36 views

Hard time factoring Normal Distribution based on transformation problem.

My professor gave problems out to practice for our final on Wednesday. This problem is based on the transformation of two random variables. It a 5 part problem, so I will list the necessary portions ...
3
votes
3answers
58 views

What is the sum of the prime factors of $2^{16}-1$?

I know $2^{10}=1024$ and $2^6=64$, but it seems they are not very helpful in solving this problem. There must be a trick to solve the problem in an easy way. What is the sum of the prime factors ...
0
votes
0answers
12 views

Odd factorisation of Indicies?

I'm sorry for the terrible formatting, I'm sorta new so I don't know how to use MathJax very well :( see below for my first attempt :) I came across the indical expression : $$\frac{x-1}{x-x^{1/2}-2}$...
0
votes
1answer
57 views

Proof that $a^{n}+b^{n}$ is irreducible over $\mathbb Q$

The sum of fourth powers cannot be factored over $\mathbb Q$, since $ a^4+b^4 = (a^2+\sqrt{2}ab+b^2)(a^2-\sqrt{2}ab+b^2)$ And these quadratic factors does not have any real rational factors. How ...
1
vote
2answers
64 views

How to factor the trinomial : $ xy-x+y-1$?

How to factor the trinomial : $ xy-x+y-1$ ? The factorization is $(x+1)(y-1) $ but I don't where it comes from.
5
votes
2answers
446 views

Is it possible to efficiently factor a semiprime given a bit-permutation relating the factors?

Is it possible to efficiently factor a semiprime given a bit-permutation relating the factors? For example, suppose we have $n = p * q = 167653$; in this case, $p = 359 = 101100111_2$ and $q = 467 = ...
1
vote
2answers
51 views

Is $X^5+…+1 \in \mathbb{F_2}[X]$ irreducible?

I am trying to determine if the following polynomials are irreducible in $\mathbb{F_2}[X]$ are irreducible: $f(X)=X^5+X^2+1$ $g(X)=X^5+X^3+1$ There are no linear factors since $f(0)=f(1)=g(...
22
votes
0answers
334 views

Irreducibility of $~\frac{x^{6k+2}-x+1}{x^2-x+1}~$ over $\mathbb Q[x]$

The Artin—Schreier polynomial $~x^n-x+1~$ is always irreducible over $\mathbb Q[x]$, unless $n=6k+2$, in which case it seems to have only two factors, one of which is always $x^2-x+1$. The ...
1
vote
0answers
21 views

Analog of Euler's Factorization method

One of Euler's discoveries was if an integer $n$ can be represented as a sum of two squares in two distinct ways, then one can factor $n$ explicitly. Of course, the method was ineffective as an ...
12
votes
0answers
173 views

Have I discovered an analytic function allowing quick factorization?

So I have this apparently smooth, parametrized function: The function has a single parameter $ m $ and approaches infinity at every $x$ that divides $m$. It is then defined for real $x$ apart ...
0
votes
1answer
23 views

Factor out $(m+2)$ in the following equation $(m+1)(m+2)+2(m+2)$

$(m+1)(m+2)+2(m+2)$ I really needed hints here, I am thinking to start at first two paragraphs and so on. Is my thought correct? Hints will be much appreciated.
3
votes
3answers
207 views

The ultimate formula to factor them all.

Context I am working on Integer factorization problem, I found a formula for factoring numbers, and I need your help to simplify it. First I will explain how I get there and then I present the ...
1
vote
2answers
33 views

Finding value of $m$ such that such that the polynomial is factorized

A polynomial $2x^2+mxy+3y^2-5y-2$ Find the value of $m$ much that $p(xy)$ can be factorized into two linear factors
1
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2answers
41 views

Stuck on a simple factoring problem

The answer to this question is probably very obvious but I can't figure it out for some reason: I simply want to factorise: $x^2+5x-2$ I solve $x^2+5x-2 = 0$ i find $x_1 = \dfrac{-5-\sqrt{33}}{2}$ ...
0
votes
1answer
31 views

Factor polynomial into linear factors with complex coefficients.

Question: A polynomial is given. $(a)$ Factor it into linear and irreducible quadratic factors with real coefficients. $(b)$ Factor it completely into linear factors with complex coefficients. $x^3 - ...
-4
votes
3answers
46 views

Factoring $12e^{2x} - 32e^x + 16$ [closed]

Can you help me solve the quadratic equation $12e^{2x}-32e^x+16$ by factoring please?
1
vote
1answer
25 views

Probability that a random polynomial over a finite field can be factorized to linear terms.

Suppose that $f\in\mathbb{F}_p[x]$ is a degree $d$ random univariate polynomial with coefficients from a finite field $\mathbb{F}_p$. What is the probability that $f$ can be written as: $$f(x)=\prod_{...
2
votes
1answer
44 views

Under what conditions on $f$, is $f(az)=g(a)f(z)$?

Formal Statement Given nonzero constant $a \in \mathbb{C}$, $|a|>0$ and $f:\mathbb{C} \to \mathbb{C}$, under what conditions on $f$ does the following hold? \begin{equation} f\left(a z\...
0
votes
0answers
13 views

On factoring given $PQ-1$ has small factors.

Suppose we have an RSA number $PQ$ where $PQ-1$ has small factors. Will this give any advantage to factor $PQ$?
1
vote
2answers
66 views

Factoring a degree 4 polynomial without power of 2 term

For my hobby, I'm trying to solve $x$ for $ax^4 + bx^3 + dx + e = 0$. (note there's no $x^2$) I hope there is a simple solution. I'm trying to write it as $(fx + g)(hx^3+i) = 0$ It follows that $fh=...
0
votes
1answer
31 views

Notation for separating out factors of a number

I have an integer (let's call it $n$), and I want to define it as the product of two values: one that's a pure power of two, and another that is odd. Obviously, these two values are unique for a ...
1
vote
2answers
38 views

Factor out (m+1) in the following so that the final answer is $\frac{(2m+1) (m+2) (m+1)} {6}$

Question: $\frac{m (m+1) (2m+1) + 6(m+1)^2}{6}$=$\frac{(2m+3)(m+2)(m+1)}{6}$ I must multiply by 6 on both sides and expand the brackets and collect like terms. I'm I correct? Edit notes: The ...
1
vote
1answer
47 views

Factor Completely.

Again, this question if from my final practice exam. Factor Completely. $$81x^4-256y^4$$ I'm able to get this far, How do I know which of the two factors should be factored further. $$(9x^2+16y^2)(...