Questions tagged [factoring]

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

If an element $n$ of the integers or more generally of a ring, e.g. a ring of polynomials, can be written as $n=ab$ with $a,b$ in the same ring, then $a,b$ are factors or divisors of $n$. If one of the factors is a unit (a divisor of 1), the factorization is called trivial. Finding a non-trivial factorization (if one exists) can be a daunting task and is the aim of many algorithms. A good factorization method might even render some strong encrpytion methods obsolete.

Having such a factorization can be very helpful, e.g. to simplify tasks: Finding the roots of $x^5-3x^2-5x^2+15$ may be difficult without knowing the factorization $(x^2-3)(x^3-5)$.

The (factoring) tag is suitable for questions about finding factors of a specific number, polynomial etc. or about aspects of general factorization methods or the structure of the set of factors.

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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 that the original limit is equal to the new and…
Emi Matro
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What is a real world application of polynomial factoring?

The wife and I are sitting here on a Saturday night doing some algebra homework. We're factoring polynomials and had the same thought at the same time: when will we use this? I feel a bit silly because it always bugged me when people asked that in…
Dan
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Fermat's Last Theorem and Kummer's Objection

In 1847 Lamé had announced that he had proven Fermat's Last Theorem. This "proof" was based on the unique factorization in $\mathbb{Z}[e^{2\pi i/p}]$. However, Kummer, proved that when $p=23$ we do not have unique factorization, and in fact Kummer…
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Could G. H. Hardy make a product of two primes so big he couldn't find out which?

This question of course began as a slightly irreverent play on the riddle, "Can God make a stone so big He can't lift it?" Then I wondered about the answer. Is it possible to exhibit a number that is provably a product of exactly two primes without…
minopret
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Is factoring polynomials as hard as factoring integers?

There seems to be a consensus that factorization of integers is hard (in some precise computational sense.) Is it known whether polynomial factorization is computationally easy or hard? One thing I originally thought was that if we could factor…
Chris Brooks
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Is there a simple explanation why degree 5 polynomials (and up) are unsolvable?

We can solve (get some kind of answer) equations like: $$ ax^2 + bx + c=0$$ $$ax^3 + bx^2 + cx + d=0$$ $$ax^4 + bx^3 + cx^2 + dx + e=0$$ But why is there no formula for an equation like $$ax^5 + bx^4 + cx^3 + dx^2 + ex + f=0$$ I'm not sure if this…
Pacerier
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Solve $x^4+3x^3+6x+4=0$... easier way?

So I was playing around with solving polynomials last night and realized that I had no idea how to solve a polynomial with no rational roots, such as $$x^4+3x^3+6x+4=0$$ Using the rational roots test, the possible roots are $\pm1, \pm2, \pm4$, but…
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Factorize $(x+1)(x+2)(x+3)(x+6)- 3x^2$

I'm preparing for an exam and was solving a few sample questions when I got this question - Factorize : $$(x+1)(x+2)(x+3)(x+6)- 3x^2$$ I don't really know where to start, but I expanded everything to get : $$x^4 + 12x^3 + 44x^2 + 72x + 36$$ I used…
A Googler
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Irreducibility of $x^n-x-1$ over $\mathbb Q$

I want to prove that $p(x):=x^n-x-1 \in \mathbb Q[x]$ for $n\ge 2$ is irreducible. My attempt. GCD of coefficients is $1$, $\mathbb Q$ is the field of fractions of $\mathbb Z$, and $\mathbb Z$ is UFD. Hence, $p(x)$ is irreducible over $\mathbb Q$…
Canis Lupus
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The coefficients of a product of monic polynomials are $0$ and $1$; if the polynomials' coefficients are non-negative, must they also be $0$ and $1$?

I have been stuck with this question for… quite sometime. Implications are mentioned after my question. Is it true that if two real polynomials $P(x), Q(x)$, have their product equal to a 0-1 polynomial (e.g. $1+x+x^5+x^{42}$), and their…
Emmanuel Amiot
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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 from a finite set of arguments. The animation above…
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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 irreducibility of its other factor,…
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How to factor a fourth degree polynomial

I'm working on a math problem but I am having a hard time figuring out the method used by my textbook to make this factorization: $$x^4 + 10x^3 + 39x^2 + 70x + 50 = (x^2 + 4x + 5)(x^2 + 6x + 10)$$ I've tried to see if this equation can be factored…
Loop
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What is the sum of all positive even divisors of 1000?

I know similar questions and answers have been posted here, but I don't understand the answers. Can anyone show me how to solve this problem in a simple way? This is a math problem for 8th grade students.Thank you very much! What is the sum of all…
learning
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Derivation of factorization of $a^n-b^n$

How does one prove that: $$a^n-b^n=(a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\dots+a^2b^{n-3}+ab^{n-2}+b^{n-1}\right)$$ Better yet, why is $a^n-b^n$ divisible by $a-b$? I would very much appreciate some help on this. Thanks
TrueDefault
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