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factorize

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Factoring Calculator - MathPapa
Shows you step-by-step how to factor expressions! This calculator will solve your problems. www.mathpapa.com
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Factorization - Wikipedia
Factoring consists of writing a number or another mathematical object as a product of several factors, usually smaller or simpler objects of the same kind. en.wikipedia.org
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Factoring Calculator - Symbolab
Factoring is a fundamental mathematical technique wherein smaller components—that is, factors—help to simplify numbers or algebraic expressions. This method ... www.symbolab.com
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factorize
factorize, v. (ˈfæktəraɪz) [f. as factorial a.1 + -ize.] 1. trans. (U.S. Law.) In Vermont and Connecticut, = garnish.1864 in Webster. 1878 [see factor n. 5 c.]. 2. Math. a. To break up (a quantity) into factors.1886 G. Chrystal Algebra I. vii. 133 The decomposition of x2 + y2 + xy = (x + y + √xy ) (... Oxford English Dictionary
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FACTORIZE | definition in the Cambridge English Dictionary
If you factorize a number, you divide it into factors. SMART Vocabulary: related words and phrases Addition, subtraction, multiplication & division dictionary.cambridge.org
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Integer factorization calculator - Alpertron
This web application factors numbers or numeric expressions using two fast algorithms: the Elliptic Curve Method (ECM) and the Self-Initializing Quadratic ... www.alpertron.com.ar
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factorize
factorizefactorise, / ˈfæktəraɪz; `fæktəˌraɪz/ v [Tn](mathematics 数) find the factors of (a number) 分解（某数的）因数. 牛津英汉双解词典
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Factoring Calculator - Free Math Help
Free online factoring calculator that factors an algebraic expression. Enter a polynomial, or even just a number, to see its factors. www.freemathhelp.com
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Factoring Calculator - Calculator Soup
The Factoring Calculator finds the factors and factor pairs of a positive or negative number. Enter an integer number to find its factors. www.calculatorsoup.com
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Factorization of Algebraic Expressions| Identities |Examples - Cuemath
Factorization of algebraic expressions is a method of finding factors for any algebraic expression. Learn more about the factorization of algebraic ... www.cuemath.com
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FACTORIZE Definition & Meaning - Dictionary.com
Mathematics., to resolve into factors. Law., garnishee. factorize. / ˈfæktəˌraɪz /. verb. (tr) maths to resolve (an integer or polynomial) ... www.dictionary.com
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Factorize x^3+3 Task: Factorize $x^3+3$ in $\mathbb{R}$ and $GF(7)$ I think the solution is $x^3+3$ in both cases, so the polynomial already is irreductible. Is my assuption is right, how do i show that?
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...
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Factorize $8x^3 + 12x^2 -2x -3$ How do I factorize this - $$8x^3 + 12x^2 -2x -3$$ I tried splitting the middle term but that didn't work , I tried factor theorem with various factors but even that didn't work. What ca...
One possibility is to note the $8x^3$ and $3$ at each end and then, thinking of the rational root theorem, see if any of $x=\pm3,\pm1,\pm\frac32,\pm\frac12,\pm\frac34,\pm\frac14,\pm\frac38,\pm\frac18$ make your expression zero. Three of them do in this case, namely $x=-\frac32,+\frac12 \text{ and } ...
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Factorize $a^2-ab-bc\pm c^2$ I got this question in a test but it did not specify the variable with respect to which I was supposed to factorize $$a^2-ab-bc\pm c^2$$ where it could be just $a(a-b)-c(b\pm c)$ but no ...
We observe that \begin{eqnarray} a^2-ab-bc+c^2&=&a^2+c^2-b(a+c)\\\ &=&(a+c)^2-b(a+c)-2ac\\\ &=&(a+c)^2-b(a+c)+\frac{b^2}{4}-\frac{b^2+8ac}{4}\\\ &=&\left(a+c-\frac{b}{2}\right)^2-\frac{b^2+8ac}{4}. \end{eqnarray} Hence, if $b^2+8ac\geq 0$ then $$ a^2-ab-bc+c^2=\left(a+c-\frac{b}{2}+\sqrt{\frac{b^2+8...
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Factorize a number into coprime numbers I want to know if there is a way to factorize a number into coprime numbers; for example $N = a_1 \cdot a_2 \cdot a_3 \cdots a_i$ And $a_i$ and $a_j$ are coprime for any $i \n...
You can construct this from the prime factorization. Just group all the appearances of the same prime factor to obtain $N=\prod p_i^{n_i}$. Since $p_i,p_j$ are distinct primes, any power of them has no non-unit common factors.
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