Partial fractions are used to simplify more complex algebraic fractions into simpler components. The ability to find partial fractions is used widely in calculus, like when solving problems involving Laplace transforms in engineering, analysing frequency response in electronics, and simplifying expressions in mathematical models. Use this resource to learn to find partial fractions.
The opposite of adding two fractions is to split one more complex fraction into partial fractions. Therefore, a partial fraction is an algebraic fraction broken into simpler parts that are added together.
For example:
\[\frac{9x-7}{(3x+4)(x-5)}=\frac{3}{3x+4}+\dfrac{2}{x-5}\]
The partial fractions are \(\dfrac{3}{3x+4}\) and \(\dfrac{2}{x-5}\).
Partial fractions are particularly useful in calculus, like when the denominator consists of linear or quadratic factors. Understanding how to find partial fractions is important for Integration using partial fractions.
Finding partial fractions
To express an algebraic fraction as partial fractions:
Factorise the denominator. They may be:
distinct linear factors:
\[\dfrac{mx+k}{(x+a)(x+b)}=\dfrac{A}{x+a}+\dfrac{B}{x+b}\]
repeated linear factors:
\[\dfrac{mx+k}{(x+a)^{2}}=\dfrac{A}{x+a}+\dfrac{B}{(x+a)^{2}}\]
quadratic and linear factors:
\[\dfrac{mx+k}{(ax^{2}+bx+c)(px+q)}=\dfrac{Ax+B}{ax^{2}+bx+c}+\dfrac{C}{px+q}\]
Write the algebraic fraction as the sum of partial fractions with unknown constants (e.g. \(A\), \(B\), \(C\)).
Add the partial fractions by getting a common denominator.
Equate the coefficients and use simultaneous equations, or substitute a value of \(x\) to determine the values of the constants.
Conditions for partial fractions
You can ONLY find partial fractions when the degree of the numerator is less than the degree of the denominator—that is, the greatest power of \(x\) on the top must be less than the greatest power of \(x\) on the bottom.
You may need to divide the denominator into the numerator and then express the remaining fractional part as partial fractions.
Express \(\dfrac{x-5}{x^{2}+2x-3}\) as a sum of partial fractions.
First, we factorise the denominator.
\[\frac{x-5}{x^{2}+2x-3}=\frac{x-5}{(x-1)(x+3)}\]
Here, we have distinct linear factors. We can separate them out with constants \(A\) and \(B\) to find the generalised partial fraction form:
\[\begin{align*} \frac{x-5}{x^{2}+2x-3} & = \frac{x-5}{(x-1)(x+3)}\\
& = \frac{A}{x-1}+\frac{B}{x+3}
\end{align*}\]
Next, we combine them again by cross-multiplying.
\[\begin{align*} \frac{A}{x-1}+\frac{B}{x+3} & = \frac{A(x+3)}{(x-1)(x+3)}+\frac{B(x-1)}{(x-1)(x+3)}\\
& = \frac{A(x+3)+B(x-1)}{(x-1)(x+3)}
\end{align*}\]
We can equate the numerator of the original fraction with our new numerator written in terms of \(A\) and \(B\).
\[\begin{align*} x-5 & = A(x+3)+B(x-1)\\
& = Ax+3A+Bx-B\\
& = (A+B)x+(3A-B)
\end{align*}\]
Again, we can equate the terms involving \(x\) and the constants on the left- and right-hand side.
\[\begin{align*} x & = (A+B)x\\
1 & = A+B\textrm{assuming }x\neq0
\end{align*}\]\[-5 = 3A-B\]
We have two simultaneous equations that we can solve. We will look at just one method of solving them.
First, let's add the two equations.
\[\begin{align*} -4 & = 4A\\
4A & = -4\\
A & = -1
\end{align*}\]
Substituting this result in the second equation:
\[\begin{align*} 1 & = -1+B\\
B & = 2
\end{align*}\]
We can then substitute \(A=-1\) and \(B=2\) into the generalised partial fraction form:
\[\frac{x-5}{x^{2}+2x-3}=\frac{-1}{x-1}+\frac{2}{x+3}\]
Express \(\dfrac{5x^{2}+3x+1}{x^{3}-3x-2}\) as the sum of partial fractions.
\[\begin{align*} \frac{5x^{2}+3x+1}{x^{3}-3x-2} & =\frac{5x^{2}+3x+1}{(x+1)^{2}(x-2)}\\
& = \frac{A}{x+1}+\dfrac{B}{(x+1)^{2}}+\frac{C}{x-2}\\
& = \dfrac{A(x+1)(x-2)+B(x-2)+C(x+1)^{2}}{(x+1)^{2}(x-2)}\\
5x^{2}+3x+1 & = A(x+1)(x-2)+B(x-2)+C(x+1)^{2}
\end{align*}\]
We can now substitute any convenient value for \(x\). This is another method for solving for \(A\), \(B\) and \(C\).
First, let \(x=2\). This lets us eliminate \(x\), \(A\) and \(B\) so that we can solve for \(C\).
\[\begin{align*} 5(2)^{2}+3(2)+1 & = A(2+1)(2-2)+B(2-2)+C(2+1)^{2}\\
27 & = A(0)+B(0)+9C\\
27 & = 9C\\
C & = 3
\end{align*}\]
To generate a second equation, let \(x=-1\). This lets us eliminate \(x\), \(B\) and \(C\) so that we can solve for \(A\).
\[\begin{align*} 5(-1)^{2}+3(-1)+1 & = A(-1+1)(-1-2)+B(-1-2)+C(-1+1)^{2}\\
3 & = A(0)+B(-3)+C(0)\\
3 & = -3B\\
B & = -1
\end{align*}\]
Substitute in the values \(C=3\) and \(B=-1\) and let \(x=0\). This lets us find \(A\).
\[\begin{align*} 5(0)^{2}+3(0)+1 & = A(0+1)(0-2)+(-1)(0-2)+(3)(0+1)^{2}\\
1 & = -2A+2+3\\
1-2-3 & = -2A\\
-4 & = -2A\\
A & =2
\end{align*}\]
The final answer is:
\[\frac{5x^{2}+3x+1}{(x+1)^{2}(x-2)} = \frac{2}{x+1}-\dfrac{1}{(x+1)^{2}}+\frac{3}{x-2}\]
Express \(\dfrac{x^{2}-3x+16}{x^{3}-5x^{2}+x-5}\) as a sum of partial fractions.
\[\begin{align*} \frac{x^{2}-3x+16}{x^{3}-5x^{2}+x-5} & = \frac{x^{2}-3x+16}{x^{2}(x-5)+(x-5)}\\
& = \frac{x^{2}-3x+16}{(x^{2}+1)(x-5)}\\
& = \frac{Ax+B}{x^{2}+1}+\frac{C}{x-5}\\
& = \frac{(Ax+B)(x-5)}{(x^{2}+1)(x-5)}+\frac{C(x^{2}+1)}{(x-5)(x^{2}+1)}\\
x^{2}-3x+16 & = (Ax+B)(x-5)+C(x^{2}+1)\\
& = Ax^{2}-5Ax+Bx-5B+Cx^{2}+C\\
& = (A+C)x^{2}+(-5A+B)x+(-5B+C)
\end{align*}\]
Equating terms on the left- and right-hand side, we have:
\[\begin{align*} \textrm{for }x^{2}\quad 1 & = A+C\\
\textrm{for }x\quad-3 & = -5A+B\\
\textrm{for }x^{0}=\textrm{constant}\quad16 & = -5B+C
\end{align*}\]
We need to solve these simultaneously. Here, we will multiply the first equation by \(5\) and add it to the second equation to find \(B\).
\[\begin{align*} 5-3 & = 5A+5C+-5A+B\\
2 & = 5C+B\\
B & = 2-5C
\end{align*}\]
Substitute this in the last equation to find \(C\).
\[\begin{align*} 16 & = -5(2-5C)+C\\
& = -10+25C+C\\
& = -10+26C\\
26C & = 26\\
C & = 1
\end{align*}\]
Substitute \(C=1\) in the first equation to find \(B\).
\[\begin{align*} 1 & = A+1\\
A & = 0
\end{align*}\]
Substitute \(A=0\) in the second equation.
\[\begin{align*} -3 & = B\\
B & = -3
\end{align*}\]
Finally, we get our solution by substituting \(A=0\), \(B=-3\) and \(C=1\) into the generalised partial fraction form:
\[\frac{x^{2}-3x+16}{x^{3}-5x^{2}+x-5}=\frac{-3}{x^{2}+1}+\frac{1}{x-5}\]
Exercise – finding partial fractions
Rewrite each of the following in the appropriate generalised partial fraction form. You do not need to calculate the constants.
