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I have begun work on Partial fractions and I've come across a problem. I would like to know if you have an irreducible quadratic term in the denominator why is the numerator $Bx + C$ instead of the usual $B$

Take this example: $\frac{x-3}{x^3+3x} \equiv \frac{x-3}{x(x^2+3)} $ Why would the partial fractions of $ \frac{x-3}{x(x^2+3)}$ take this form $\frac{A}{x} + \frac{Bx + C}{x^2+3}$

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  • $\begingroup$ Because $A$ is multiplied by $x^2+3$ and hence we would have a $x^2$ in the nominator. How can we cancel that term? $\endgroup$ – polfosol Oct 29 '16 at 6:10
  • $\begingroup$ @polfosol wouldn't A = 0, since there are no $x^2$ terms in the original fraction? $\endgroup$ – user383805 Oct 29 '16 at 6:14
  • $\begingroup$ Now put $A=0$ to see what you finally get $\endgroup$ – polfosol Oct 29 '16 at 6:15
  • $\begingroup$ @polfosol lol sorry silly question $\endgroup$ – user383805 Oct 29 '16 at 6:17
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    $\begingroup$ There is no such thing as silly question $\endgroup$ – polfosol Oct 29 '16 at 6:18
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Before we do partial fractions on a rational expression, we make sure that it's a proper rational expression. If it's not, we have to do long division (or something) to rewrite it as a polynomial plus a proper rational expression, and then do partial fractions to the proper rational expression.

Then when we find the expansion, we expect the terms to be proper also. In order to account for all possibilities, the degree of the numerator must be assumed to be one less than that of the denominator (because that make the term proper.) So with a quadratic (degree 2) denominator, we need to check for linear (degree 1) numerator. With a linear (degree 1 denominator, we need check only for constant (degree 0) numerator.

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Try and write $$ \frac{1}{x(x^2+3)} = \frac{A}{x} + \frac{B}{x^2+3}, $$ with $A, B$ constants. You get $$ A (x^{2} + 3) + B x = 1, $$ which clearly has no solutions $A, B$. This is because $A x^{2} + B x + 3 A = 1$ implies $A = 0$ and $0 = 3 A = 1$, a contradiction.


On the other hand, if $f, g$ are two coprime non-constant polynomials, there exist polynomials $a, b$ such that $$ a f + b g = 1. $$ Divide $a$ by $g$ to get $a = u g + v$, where $v$ has degree less than $g$. Then you have also $$ (a - u g) f + (b + u f) g = 1, $$ with $a - u g$ of degree less than $g$. Since $(a - u g) f$ and $(b + u f) g$ sum to $1$, they have the same degree, so that the degree of $b + u f$ is less than the degree of $f$. You have written $$ a'f + b'g = 1, $$ that is, $$ \frac{1}{f g} = \frac{a'}{g} + \frac{b'}{f} $$ with $a'$ of degree less than $g$ and $b'$ of degree less than $f$.

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