# How to express $\frac{x^3+4x^2-1}{(x^2+1)^2}$ as a polynomial plus a proper fraction, using long division?

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.

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What 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? – Arturo Magidin Mar 27 '12 at 22:16
You aren't, by any chance, trying to do "partial fractions" and write it as $(Ax+B)/(x^2+1)+(Cx+D)/(x^2+1)^2$? – Gerry Myerson Mar 27 '12 at 22:20
@ArturoMagidin I mean factorize using the polinomial large divisions. – Garmen1778 Mar 27 '12 at 22:21
@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. – Arturo Magidin Mar 27 '12 at 22:22
@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. – Arturo Magidin Mar 27 '12 at 22:27

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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What is the $deg(r)$ function? – Garmen1778 Mar 27 '12 at 23:18
@Garmen1778: it is the polynomial degree, the exponent of the highest power of the independent variable that has non-zero coefficient; e.g. $\deg(x^3+4x^2-1)=3$ – robjohn Mar 27 '12 at 23:46

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.

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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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see the comments; you are answering the only reasonable reading of the question, but not the intended question... – Arturo Magidin Mar 27 '12 at 22:35