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For example I have $f(x)=x+1$ and $g(x)=x^2+2x+1$:

I don't have any problem when I calculate $g(x)/f(x)$, but I tried to calculate $f(x)/g(x)$ and I couldn't find the way to solve it.

  1. How should I do the polynomial long division for that situation?

  2. If I have an expression $\frac{1}{a+b}$, can I divide it into two different terms?

abouttostart
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  • Hint $g(x) = (x + 1)^2$. – Travis Willse Jun 03 '17 at 11:19
  • @Travis I could solve that actual problem by that way, but I mean if it cannot be factored, I was asking for comprehensive ways of division. – abouttostart Jun 03 '17 at 13:38
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    It depends on what you mean. In polynomial long division (for polynomials over, say, $\Bbb Z$ or a field), if divide $f(x)$ by $g(x)$, we effectively find (unique) polynomials $q(x)$ and $r(x)$ such that $f(x) = q(x) g(x) + r(x)$, where the remainder $r$ satisfies $\deg r < \deg g$. If $\deg f < \deg g$, then the only way to write $f$ this way is as $f(x) = 0 \cdot g(x) + f(x)$. Put another way, if $\deg f < \deg g$, in this sense there's nothing to do. – Travis Willse Jun 03 '17 at 14:30

2 Answers2

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Since $f(x)$ has a lower degree, the quotient will be $0$ and the remainder will be $f(x)$.

$$f(x) = 0g(x) + (x+1)$$

There is no way that the quotient $q(x)$ be something non-zero (constant or higher degree), that would make $q(x)g(x)$ something with degree 1 or less.

peterwhy
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When you divid a polynomial $f(x)$ by another polynomial $g(x)$ whose degree is greater, the quotient is $0$ and the remainder is $f(x)$. That is, $f(x)=0\times g(x)+f(x)$.

In your specific situation, you get:$$\frac{f(x)}{g(x)}=\frac{x+1}{(x+1)^2}=\frac1{x+1}.$$

José Carlos Santos
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