# Why, in partial fraction decomposition, does an irreducible quadratic term in the denominator require a linear term in the numerator?

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 $\frac{A}{x} + \frac{Bx + C}{x^2+3}$

• 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? – polfosol Oct 29 '16 at 6:10
• @polfosol wouldn't A = 0, since there are no $x^2$ terms in the original fraction? – user383805 Oct 29 '16 at 6:14
• Now put $A=0$ to see what you finally get – polfosol Oct 29 '16 at 6:15
• @polfosol lol sorry silly question – user383805 Oct 29 '16 at 6:17
• There is no such thing as silly question – polfosol Oct 29 '16 at 6:18

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 $f$.