Topic 10 Partial fractions
A rational function is a quotient of two polynomials, i.e. an expression of the form \(\dfrac{p_1(x)}{p_2(x)}\), where both \(p_1(x)\) and \(p_2(x)\) are polynomials, and \(p_2(x)\neq0\).
For example: \(\dfrac{5x+7}{3x^2+5}\), \(\dfrac{5x^2-4x+3}{x-7}\), …
Long division of polynomials allows us to write \(\dfrac{p_1(x)}{p_2(x)}\) in the form \[ \frac{p_1(x)}{p_2(x)} = q(x) + \frac{r(x)}{p_2(x)}, \] where \(q(x)\) and \(r(x)\) are polynomials and the degree of \(r(x)\) is less than the degree of \(p_2(x)\).
The aim of “partial fractions”: simplify \(\dfrac{r(x)}{p_2(x)}\) further.
First factorise the denominator \(p_2(x)\) (completely1).
Then, every rational function of the type as in the left–hand column can be split into partial fractions as given in the right–hand column:
| Type I | \(\dfrac{Px+Q}{(ax+b)(cx+d)}\) | = | \(\dfrac{A}{ax+b}+\dfrac{B}{cx+d}\) |
| Type II | \(\dfrac{Px+Q}{(ax+b)^2}\) | = | \(\dfrac{A}{ax+b}+\dfrac{B}{(ax+b)^2}\) |
| Type III | \(\dfrac{Px^2+Qx+R}{(ax+b)^2(cx+d)}\) | = | \(\dfrac{A}{ax+b}+\dfrac{B}{(ax+b)^2}+\dfrac{C}{cx+d}\) |
| Type IV | \(\dfrac{Px^2+Qx+R}{(ax^2+bx+c)(dx+e)}\) | = | \(\dfrac{Ax+B}{ax^2+bx+c}+\dfrac{C}{dx+e}\) |
Note: It is a Theorem that the above actually works — these are equations which work for all values of \(x\) simultaneously! It is not proved here though.
Example 10.1 Split \(\dfrac{-4x+15}{(x+2)(x-1)}\) into partial fractions.
Solution. First, note that the degree of the numerator is less than the degree of the denominator, so we do not need to perform any long division.
Second, the denominator is already factored as much as possible (into linear factors). We see that it is “Type I”, so we want to find \(A\) and \(B\) such that
\[
\frac{-4x+15}{(x+2)(x-1)} = \frac{A}{x+2}+\frac{B}{x-1}
\]
Multiplying both sides by \((x+2)(x-1)\), we arrive at
\(-4x+15 = A(x-1)+B(x+2)\).
Let \(x=1\). Then \(-4+15 = A\cdot 0+B\cdot 3\) \(\implies\) \(B=\frac{11}{3}\).
Let \(x=-2\). Then \(8+15=A(-3)+B\cdot 0\) \(\implies\) \(A=-\frac{23}{3}\).
Answer:
\(\dfrac{-4x+15}{(x+2)(x-1)} = -\dfrac{23}{3(x+2)}+\dfrac{11}{3(x-1)}.\)
Example 10.2 Split \(\dfrac{5x^2-3x+2}{(x-3)^2(x+2)}\) into partial fractions.
Solution. (Type III.) We want to find \(A,B,C\) such that
\[
\frac{5x^2-3x+2}{(x-3)^2(x+2)}=\frac{A}{x-3}+\frac{B}{(x-3)^2}+\frac{C}{x+2}.
\]
\(\iff\) \(5x^2-3x+2 = A(x-3)(x+2)+B(x+2)+C(x-3)^2\).
Let \(x=-2\). Then \(5\cdot 4-3(-2)+2 = C\cdot(-5)^2\) \(\implies\) \(C=\frac{28}{25}\).
Let \(x=3\). Then \(5\cdot 9-3\cdot 3+2 = B\cdot 5\) \(\implies\) \(B=\frac{38}{5}\).
Equating coefficients of \(x^2\) gives \(5=A+C\) \(\implies\) \(A=5-C=5-\frac{28}{25}=\frac{97}{25}\).
Answer:
\(\dfrac{5x^2-3x+2}{(x-3)^2(x+2)}=\dfrac{97}{25(x-3)}+\dfrac{38}{5(x-3)^2}+\dfrac{28}{25(x+2)}.\)
Example 10.3 Split \(\dfrac{2x+5}{(x^2+x+1)(x+3)}\) into partial fractions.
Solution. (Type IV.) We want to find \(A,B,C\) such that
\[\dfrac{2x+5}{(x^2+x+1)(x+3)}=\dfrac{Ax+B}{x^2+x+1}+\dfrac{C}{x+3}.\]
\(\iff\) \(2x+5=(Ax+B)(x+3)+C(x^2+x+1)\).
Let \(x=-3\). Then \(2\cdot(-3)+5=C\cdot(9-3+1)\) \(\implies\) \(C=-\frac17\).
Equating coefficients of \(x^2\) gives \(0=A+C\) \(\implies\) \(A=\frac17\).
Equating constant coefficients gives \(5=3B+C\) \(\implies\) \(B=\frac{5-C}{3}=\frac{12}7\).
Answer:
\(\dfrac{2x+5}{(x^2+x+1)(x+3)}=\dfrac{x+12}{7(x^2+x+1)}-\dfrac{1}{7(x+3)}.\)
When the coefficients of a polynomial are real numbers, it is always possible to factorise it into a product of factors, each of which is either linear, or an irreducible quadratic. (This is a consequence of the Fundamental Theorem of Algebra.)↩︎