The Method of Partial Fractions


2. The polynomial x2+3x+6 is irreducible, thus the setup looks like this:

\begin{displaymath}\frac{5x^2+12x+16}{(x^2+3x+6)(x+2)}=\frac{Ax+B}{x^2+3x+6}+\frac{C}{x+2}.\end{displaymath}

Multiplying by the least common denominator and "sorting" results in

5x2+12x+16=(2A+C)x2+(A+B+3C)x+(2B+6C).

By comparing the polynomial coefficients on both sides we obtain the requirements

\begin{eqnarray*}5&=&A+C\\
12&=&2A+B+3C\\
16&=&2B+6C
\end{eqnarray*}


The coefficients turn out to be A=2, B=-1 and C=3, so

\begin{displaymath}\frac{5x^2+12x+16}{(x^2+3x+6)(x+2)}=\frac{2x-1}{x^2+3x+6}+\frac{3}{x+2}.\end{displaymath}


Helmut Knaust
Sat Oct 24 13:54:22 MDT 1998

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