I'm trying to express $$\dfrac{x^3+4x^2-1}{(x^2+1)^2}$$ as a polynomial plus a proper fraction, using long division but I don't know how to do that. It'd be cool if you can solve this. Thanks.
How to express $\frac{x^3+4x^2-1}{(x^2+1)^2}$ as a polynomial plus a proper fraction, using long division?
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0What do you mean, "factor"? The fraction is already in least terms. Do you mean, factor the numerator? It does not have any rational roots, so any factorization would involve nonrational reals. Is that what you want? – 2012-03-27
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1You aren't, by any chance, trying to do "partial fractions" and write it as $(Ax+B)/(x^2+1)+(Cx+D)/(x^2+1)^2$? – 2012-03-27
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0@ArturoMagidin I mean factorize using the polinomial large divisions. – 2012-03-27
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1@Garmen1778: You mean long division? It's already in that form, since the degree of the numerator is smaller than the degree of the denominator. Sorry, but you aren't making much sense to me. Do you have an example where you *do* know the answer that you could type, so we can see what you mean? Clearly, a lot is being lost in translation. – 2012-03-27
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0@ArturoMagidin So, if the degree of the numerator is smaller than the degree of the denominator it can't be factorized? That was the problem. Answered. – 2012-03-27
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1@Garmen1778: Expressing a fraction of polynomials as a quotient times a remainder is **not** called "factorizing." Again: I don't know what you mean, and it is clear that you are trying to translate from some other language and are getting the technical terms wrong (such as "large divisiong"). So, I would *strongly urge you* to give a worked out example of what it is you are trying to do so that we can (i) tell you what is the proper way of saying it in English; and (ii) answer your question here. – 2012-03-27
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0@ArturoMagidin I was trying to do the Polynomial large division(http://en.wikipedia.org/wiki/Polynomial_long_division) and see the answer as a form of $f(x)=(mx+n)+\frac{1}{x}$ instead of $f(x)=\dfrac{mx^2+nx+1}{x}$. I'm not quite sure if you understand me… – 2012-03-27
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0@Garmen1778: That is not called "factorizing", and it's not called "large division." What you want to do is "express as a polynomial plus a proper fraction, using **long** division." (You link to Wikipedia's article that calls it "long division", but you still refer to it as **large** division; large$\neq$long) The fraction is "proper" when the degree of the numerator is strictly smaller than the degree of the denominator, and that is already the case here. – 2012-03-27
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0@Garmen1778 Please don't completely change your question (with no notice), since now you've invalidated some answers (such as mine), which may confuse readers, spur downvotes, etc. – 2012-03-27
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0@BillDubuque I don't know which person I have to ask. Should I close the question? – 2012-03-27
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0@Garmen1778 No need to close, just keep the original problem too so the prior answers make sense. Better to append text to the original problem elaborating... – 2012-03-27
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0@BillDubuque Ok thanks. – 2012-03-27
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0@Garmen1778 It's ok to mention your second related problem. Simply say something like: "Update: after reading the comments I realized .... so this this is the problem I need help with....". – 2012-03-27
3 Answers
If $f(x)$ and $g(x)$ are two polynomials, $g(x)\neq 0$, then there exist unique polynomial $q(x)$ and $r(x)$, called the "quotient" and the "remainder", such that $$\frac{f(x)}{g(x)} = q(x) + \frac{r(x)}{g(x)},\qquad r(x)=0\text{ or }\deg(r)\lt\deg(g).$$ Both $q$ and $r$ can be found using polynomial long division.
(We often refer to a fraction of polynomials in which the degree of the numerator is strictly smaller than the degree of the denominator as a "proper fraction", in analogy to the case of numerical fractions $\frac{a}{b}$, which are "proper" when $|a|\lt |b|$, and improper if $|a|\geq|b|$; every fraction $\frac{a}{b}$ with $a$ and $b$ integers, $b\neq 0$, can be written uniquely as $\frac{a}{b} = n + \frac{r}{b}$, where $n$ and $r$ are integers and $0\leq r\lt|b|$; this is the analogous operation with polynomials).
In your case, since $f(x) = x^3 + 4x -1$ and $g(x) = (x^2+1)^2$, we already have $\deg(f)\lt \deg(g)$, so the fraction is already proper; that means that $q(x) = 0$ and $r(x) = f(x)$, and there is nothing left to do.
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0What is the $deg(r)$ function? – 2012-03-27
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1@Garmen1778: it is the polynomial degree, the exponent of the highest power of the independent variable that has non-zero coefficient; e.g. $\deg(x^3+4x^2-1)=3$ – 2012-03-27
Hint $\rm\ \ x^3 + 4\: x^2 - 1\ =\ (x+4)\:(x^2+1) - (x+5)\ $ by the Polynomial division algorithm.
Now divide both sides by $\rm\:(x^2+1)^2\:$ and cancel a factor of $\rm\:x^2+1$ from one term.
By the rational root theorem, you know the only possible rational roots would be $\pm1$, neither of which is in fact a root.
The numerator factors (approximately) as $$ (x+3.93543233197003)(x+0.537401577025226)(x-0.472833908995256) $$ which are irrational algebraic roots.
You could use the cubic equation to find the closed form expression for these.
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0see the comments; you are answering the only reasonable reading of the question, but not the *intended* question... – 2012-03-27