Title: Default for 3.6 Partial Equations
Course: HELM Workbooks 2023-2024

Proper and improper fractions

Frequently we find that an algebraic fraction appears in the form

\[\text{algebraic fraction} = {\hbox{numerator}\over \hbox{denominator}}\]

where both numerator and denominator are polynomials. For example:

\(\displaystyle{{x^3+x^2+3x+7\over x^2+1}; \qquad {3x^2-2x+5\over x^2-7x+2}; \qquad {x\over x^4+1}.}\)

The degree of the numerator, \(n\) say, is the highest power occurring in the numerator. The degree of the denominator, \(d\) say, is the highest power occurring in the denominator.

If \(d>n\) the fraction is said to be proper; the third expression above is such an example.

If \(d\le n\) the fraction is said to be improper; the first and second expressions above are examples of this type.

Before calculating the partial fractions of an algebraic fraction it is important to decide whether the fraction is proper or improper.

The denominator of an algebraic fraction can often be factorised into a product of linear and/or irreducible quadratic factors. Before we can separate algebraic fractions into simpler (partial) fractions we need to completely factorise the denominators into linear and irreducible quadratic factors.

Linear factors are those of the form \(ax+b\); for example \(2x+7\), \(3x-2\) and \(4-x\).

Irreducible quadratic factors are those of the form \(ax^2+bx+c\) such as \(x^2+x+1\), and \(4x^2-2x+3\), which cannot be factorised into linear factors (these are real quadratics with complex roots).

Proper Fractions with Linear Factors

Firstly we describe how to calculate partial fractions for proper fractions where the denominator may be written as a product of linear factors. The steps are as follows:

- Factorise the denominator.
- Each factor will produce a partial fraction. A factor such as \(3x+2\) will produce a partial fraction of the form \(\dfrac{A}{3x+2}\) where \(A\) is an unknown constant. In general a linear factor \(ax+b\) will produce a partial fraction \(\dfrac{A}{ax+b}\). The unknown constants for each partial fraction may be different and so we will call them \(A\), \(B\), \(C\) and so on.
- Evaluate the unknown constants by equating coefficients or using specific values of \(x\).

The sum of the partial fractions is identical to the original algebraic fraction for all values of \(x\).

Key Point

A linear factor \(ax+b\) in the denominator gives rise to a single partial fraction of the form \(\dfrac{A}{ax+b}\).

The steps involved in expressing a proper fraction as partial fractions are illustrated in the following Example.

Example

Express \(\dfrac{7x+10}{ 2x^2+5x+3}\) in terms of partial fractions.

Solution

Note that this fraction is proper. The denominator is factorised to give \((2x+3)(x+1)\). Each of the linear factors produces a partial fraction. The factor \(2x+3\) produces a partial fraction of the form \(\dfrac{A}{2x+3}\) and the factor \(x+1\) produces a partial fraction \(\dfrac{B}{x+1}\), where \(A\) and \(B\) are constants which we need to find. We write

\[\dfrac{7x+10}{(2x+3)(x+1)} = \dfrac{A}{2x+3} + \dfrac{B}{x+1}\]

By multiplying both sides by \((2x+3)(x+1)\) we obtain

\[7x+10 = A(x+1) + B(2x+3)\qquad(*)\]

We may now let \(x\) take any value we choose. By an appropriate choice we can simplify the right-hand side. Let \(x=-1\) because this choice eliminates \(A\). We find \[\begin{aligned} 7(-1) + 10 &= A(0) + B(-2+3)\\ 3&=B \end{aligned}\] so that the constant \(B\) must equal 3. The constant \(A\) can be found either by substituting some other value for \(x\) or alternatively by ‘equating coefficients’. Observe that, by rearranging the right-hand side, Equation (*) can be written as

\[7x+10 = (A+2B)x + (A+3B)\]

Comparing the coefficients of \(x\) on both sides we see that \(7=A+2B\). We already know \(B=3\) and so \[\begin{aligned} 7&=A + 2(3)\\ &= A+ 6\end{aligned}\] from which \(A=1\). We can therefore write

\[\dfrac{7x+10}{2x^2+5x+3} = \dfrac{1}{2x+3} + \dfrac{3}{x+1}\]

We have succeeded in expressing the given fraction as the sum of its partial fractions. The result can always be checked by adding the fractions on the right.

Example

Express \(\dfrac{7x+10}{ 2x^2+5x+3}\) in terms of partial fractions.

Solution

Note that this fraction is proper. The denominator is factorised to give \((2x+3)(x+1)\). Each of the linear factors produces a partial fraction. The factor \(2x+3\) produces a partial fraction of the form \(\dfrac{A}{2x+3}\) and the factor \(x+1\) produces a partial fraction \(\dfrac{B}{x+1}\), where \(A\) and \(B\) are constants which we need to find. We write

\[\dfrac{7x+10}{(2x+3)(x+1)} = \dfrac{A}{2x+3} + \dfrac{B}{x+1}\]

By multiplying both sides by \((2x+3)(x+1)\) we obtain

\[7x+10 = A(x+1) + B(2x+3)…(*)\]

We may now let \(x\) take any value we choose. By an appropriate choice we can simplify the right-hand side. Let \(x=-1\) because this choice eliminates \(A\). We find \[\begin{aligned} 7(-1) + 10 &= A(0) + B(-2+3)\\ 3&=B \end{aligned}\] so that the constant \(B\) must equal 3. The constant \(A\) can be found either by substituting some other value for \(x\) or alternatively by ‘equating coefficients’. Observe that, by rearranging the right-hand side, Equation (*) can be written as

\[7x+10 = (A+2B)x + (A+3B)\]

Comparing the coefficients of \(x\) on both sides we see that \(7=A+2B\). We already know \(B=3\) and so \[\begin{aligned} 7&=A + 2(3)\\ &= A+ 6\end{aligned}\] from which \(A=1\). We can therefore write

\[\dfrac{7x+10}{2x^2+5x+3} = \dfrac{1}{2x+3} + \dfrac{3}{x+1}\]

We have succeeded in expressing the given fraction as the sum of its partial fractions. The result can always be checked by adding the fractions on the right.

Proper fractions with repeated linear factors

Sometimes a linear factor appears more than once. For example in

\(\dfrac{1}{x^2+2x+1}=\dfrac{1}{(x+1)(x+1)}\) which equals \(\dfrac{1}{(x+1)^2}\)

the factor \((x+1)\) occurs twice. We call it a repeated linear factor. The repeated linear factor \((x+1)^2\) produces two partial fractions of the form \(\dfrac{A}{x+1} + \dfrac{B}{(x+1)^2}\). In general, a repeated linear factor of the form \((ax+b)^2\) generates two partial fractions of the form

\[\dfrac{A}{ax+b} + \dfrac{B}{(ax+b)^2}\]

This is reasonable since the sum of two such fractions always gives rise to a proper fraction:

\[\dfrac{A}{ax+b} + \dfrac{B}{(ax+b)^2}=\dfrac{A(ax+b)}{(ax+b)^2}+\dfrac{B}{(ax+b)^2}= \dfrac{x(Aa)+Ab+B}{(ax+b)^2}\]

Key Point

A repeated linear factor \((ax+b)^2\) in the denominator produces two partial fractions:

\[\dfrac{A}{ax+b} + \dfrac{B}{(ax+b)^2}\]

Once again the unknown constants are found by either equating coefficients and/or substituting specific values for \(x\).

Proper fractions with quadratic factors

Sometimes when a denominator is factorised it produces a quadratic term which cannot be factorised into linear factors. One such quadratic factor is \(x^2+x+1\). This factor produces a partial fraction of the form \(\dfrac{Ax+B}{x^2+x+1}\). In general a quadratic factor of the form \(ax^2+bx+c\) produces a single partial fraction of the form \(\dfrac{Ax+B}{ax^2+bx+c}\).

Key Point

A quadratic factor \(ax^2+bx+c\) in the denominator produces a partial fraction of the form \[\dfrac{Ax+B}{ax^2+bx+c}\]

Improper fractions

When calculating the partial fractions of improper fractions an extra polynomial is added to any partial fractions that would normally arise. The added polynomial has degree \(n-d\) where \(n\) is the degree of the numerator and \(d\) is the degree of the denominator. Recall that

- a polynomial of degree 0 is a constant, \(A\) say,
- a polynomial of degree 1 has the form \(Ax+B\),
- a polynomial of degree 2 has the form \(Ax^2+Bx+C\)

and so on. If, for example, the improper fraction is such that the numerator has degree 5 and the denominator has degree 3, then \(n-d=2\), and we need to add a polynomial of the form \(Ax^2+Bx+C\).

Key Point

If a fraction is improper an additional term is included taking the form of a polynomial of degree \(n-d\), where \(n\) is the degree of the numerator and \(d\) is the degree of the denominator.

Example

Express as partial fractions

\[\dfrac{2x^2-x-2}{x+1}\]

Solution

The fraction is improper because \(n=2\), \(d=1\) and so \(d\le n\). Here \(n-d=1\), so we need to include as an extra term a polynomial of the form \(Bx+C\), in addition to the usual partial fractions. The linear term in the denominator gives rise to a partial fraction \(\dfrac{A}{x+1}\). So altogether we have

\[\dfrac{2x^2-x-2}{x+1}= \dfrac{A}{x+1}+ (Bx+C).\]

Multiplying both sides by \(x+1\) we find

\[2x^2-x-2 = A+(Bx+C)(x+1) = Bx^2 + (C+B)x + (C+A).\]

Equating coefficients of \(x^2\) gives

\[B=2.\]

Equating coefficients of \(x\) gives

\[-1=C+B\]

and so

\[C=-1-B=-3.\]

Equating the constant terms gives

\[-2=C+A\]

and so

\[A=-2-C=-2-(-3)=1.\]

Finally, we have

\[\dfrac{2x^2-x-2}{x+1}= \dfrac{1}{x+1}+2x-3.\]

Example

Express as partial fractions

\[\dfrac{2x^2-x-2}{x+1}\]

Solution

The fraction is improper because \(n=2\), \(d=1\) and so \(d\le n\). Here \(n-d=1\), so we need to include as an extra term a polynomial of the form \(Bx+C\), in addition to the usual partial fractions. The linear term in the denominator gives rise to a partial fraction \(\dfrac{A}{x+1}\). So altogther we have

\[\dfrac{2x^2-x-2}{x+1}= \dfrac{A}{x+1}+ (Bx+C)\]

Multiplying both sides by \(x+1\) we find

\[2x^2-x-2 = A+(Bx+C)(x+1) = Bx^2 + (C+B)x + (C+A)\]

Equating coefficients of \(x^2\) gives

\[B=2\].

Equating coefficients of \(x\) gives

\[-1=C+B\]

and so

\[C=-1-B=-3\].

Equating the constant terms gives

\[-2=C+A\]

and so

\[A=-2-C=-2-(-3)=1\].

Finally, we have

\[\dfrac{2x^2-x-2}{x+1}= \dfrac{1}{x+1}+2x-3\]

Exercise

For each of the following fractions state the degree of the numerator \((= n)\) and the degree of the denominator \((= d)\). Hence classify the fractions as proper or improper.

1. \(\displaystyle [value]\)
\(n =\)
\(d =\)
The function is
2. \(\displaystyle [value]\)
\(n =\)
\(d =\)
The function is
3. \(\displaystyle [value]\)
\(n =\)
\(d =\)
The function is
4. \(\displaystyle [value]\)
\(n =\)
\(d =\)
The function is
Feedback: 1. The degree of the numerator, \(n\) is \([value]\). The degree of the denominator, \(d\) is \([value]\). Because \([value]\) the fraction is \([value]\). 2. The degree of the numerator, \(n\) is \([value]\). The degree of the denominator, \(d\) is \([value]\). Because \([value]\) the fraction is \([value]\). 3. The degree of the numerator, \(n\) is \([value]\). The degree of the denominator, \(d\) is \([value]\). Because \([value]\) the fraction is \([value]\). 4. The degree of the numerator, \(n\) is \([value]\). The degree of the denominator, \(d\) is \([value]\). Because \([value]\) the fraction is \([value]\).

Exercise

Express \[[value]\] in partial fractions.

First factorise the denominator (find \([value]\) and \([value]\) so that \([value]=[value]\)):

\([value]=\)

Because there are two linear factors we write this as:

\[\dfrac{\text{numerator}}{\text{denominator}}=\dfrac{[value]}{[value]}+\dfrac{[value]}{[value]}\]

Multiply both sides by \([value]\) to obtain the equation from which to find \([value]\) and \([value]\):

\([value]=\)

Substitute an appropriate value for \([value]\) to obtain \([value]\):

Substitute \([value]=\) and get \([value]=\).

Equating coefficients of \([value]\) obtain the value of \([value]\):

\([value]=\).

Finally, write down the partial fractions::

\(\displaystyle{[value]=}\).
Feedback: Factorising the denominator we obtain \(([value])([value])\). We can see then that \(\dfrac{[value]}{[value]}=\dfrac{[value]}{([value])([value])}\). We want to express the right hand side as a sum of two fractions. This means, we need to find expessions \([value]\) and \([value]\) such that \[\dfrac{[value]}{([value])([value])}=\dfrac{[value]}{[value]}+\dfrac{[value]}{[value]}\] Multiplying both sides of the expression by \(([value])([value])\) we obtain \[\dfrac{[value]}{([value])([value])}([value])([value])=\dfrac{[value]}{[value]}([value])([value])+\dfrac{[value]}{[value]}([value])([value]).\] Cancelling out, we obtain: \[[value]=[value]([value])+[value]([value]).\] In order to obtain \([value]\), we can substitute \([value]=\displaystyle [value]\). From there we can see that \([value]=[value]\). Similarly we can obtain \([value]=[value]\). Finally, the correct fraction is: \[[value]\]

Exercise

Find the partial fractions of:

1. \([value]= \)

2. \([value]= \)

3. \([value]= \)

Exercise

Express \(\displaystyle[value]\) in partial fractions.

First factorise the denominator to obtain a factor of the form \(a[value]+b\) where \(a,b \in \mathbb{Z}\):

\([value]=\)

This is a repeated linear factor which gives rise to two partial fractions of the form:

\[\dfrac{\text{numerator}}{\text{denominator}}=\dfrac{[value]}{[value]}+\dfrac{[value]}{[value]}\]

Multiply both sides by your answer above to obtain the equation from which to find \([value]\) and \([value]\):

\([value]=\)

Now evaluate the constants \([value]\) and \([value]\) by equating coefficients:

\([value]=\) ; \([value]=\)

Finally express the answer in partial fractions:

\(\displaystyle[value]=\)
Feedback: We are given the fraction \(\displaystyle[value].\) We factorise the denominator to obtain: \([value]= [value].\) There is a repeated linear factor \(([value])\). Multiplying both sides by the factor \([value]\) gives \([value]\). Equating the \([value]\) coefficients gives \([value]=[value]\). Equating the free coefficients gives \([value]=[value]\). Plugging in the repeated factor \([value]\) and the correct values for \([value]\) and \([value]\) we obtain the partial fractions: \[ [value]= [value].\]

Exercise

Express the following in partial fractions.

1. \(\displaystyle{[value]=}\)

2. \(\displaystyle{[value]=}\)

3. \(\displaystyle{[value]=}\)
Feedback: 1. The repeated linear factor is \([value]\) and the partial fractions are \([value]\). 2. The repeated linear factor is \([value]\) and the partial fractions are \([value]\). 3. The repeated linear factor is \([value]\) and the partial fractions are \([value]\).

Exercise

Express each of the following as the sum of its partial fractions.

1. \(\dfrac{[value]}{[value]}=\)

2. \(\dfrac{[value]}{[value]}=\)
Feedback: 1. First, by multiplying both sides by \(([value])([value])\) we obtain \([value]\). By letting \([value]=[value]\) we find that the correct value of \([value]\) is \([value]\). By equating the coefficients of \([value]^2\) we find \([value]=[value]\). If we substitute any other value for \([value]\) (or equate coefficients of \([value]\)) we find \([value]=[value]\). Finally, the partial fractions are \(\displaystyle [value]\). 2. First, by multiplying both sides by \(([value])([value])\) we obtain \([value]\). By letting \([value]=[value]\) we find that the correct value of \([value]\) is \([value]\). By equating the coefficients of \([value]^2\) we find \([value]=[value]\). If we substitute any other value for \([value]\) (or equate coefficients of \([value]\)) we find \([value]=[value]\). Finally, the partial fractions are \(\displaystyle [value]\).

Exercise

Express each of the following improper fractions in terms of partial fractions.

1. \(\dfrac{[value]}{[value]}=\)
2. \(\dfrac{[value]}{[value]}=\)
3. \(\dfrac{[value]}{[value]}=\)
Feedback: 1. Notice that: \(\dfrac{[value]}{[value]} = \displaystyle [value].\) After multiplying both sides by the denominator and equating the coefficients, you will obtain the partial fractions \(\displaystyle[value]\). 2. Notice that: \(\dfrac{[value]}{[value]}= \displaystyle [value].\) After multiplying both sides by the denominator and equating the coefficients, you will obtain the partial fractions \(\displaystyle[value]\). 3. Notice that: \(\dfrac{[value]}{[value]}=\displaystyle [value].\) After multiplying both sides by the denominator and equating the coefficients, you will obtain the partial fractions \(\displaystyle[value]\).

Exercise

Find the partial fractions of:

\(\displaystyle [value]= \)
Feedback: We are given the expression \[\displaystyle [value].\] The numerator has degree \(1\) and the denominator has degree \(2\), therefore this is a proper fraction. There are \(2\) linear factors in the denominator, therefore \[\displaystyle [value] = \frac{A}{[value]} + \frac{B}{[value]} = \frac{A\cdot ([value]) + B\cdot ([value]) }{ ([value])([value])} .\] Equating the top line yields \[ [value] = A\cdot ([value]) + B\cdot ([value]).\] Then by substituting values of \(x\) we can solve for \(A, \, B\) and obtain \( A = [value]\) and \( B = [value].\) Therefore: \[ [value]= [value].\]

Exercise

Express the following in partial fractions.

\(\displaystyle{[value]=}\)
Feedback: We are given the expression \[\displaystyle{[value]}.\] The repeated linear factor is \([value]\). Therefore the partial fraction decomposition is \[ [value] = \frac{A}{[value]} + \frac{B}{([value])^2} = \frac{A\cdot ([value]) + B}{ ([value])^2}. \] Equating the top line yields \[ [value] = A\cdot ([value]) + B.\] Then by substituting values of \(x\) we can solve for \(A, \,B\) and obtain \( A = [value]\) and \(B = [value].\) Therefore \[[value] = [value].\]

Exercise

Express as partial fractions:

\[\dfrac{[value]}{([value])([value])}\]

Please pretend that the quadratic factor cannot be factorised further, it can be, but I can't change the question now. Let's imagine a brighter world where the rest of this makes sense.

In our brighter world we have

\[\dfrac{[value]}{([value])([value])}=\dfrac{[value]}{[value]}+\dfrac{[value]}{[value]}\]

First multiply both sides by \(([value])([value])\):

\([value]=\)

Evaluate \([value]\):

\([value]=\)

Equate coefficients of \([value]^2\) and hence find \([value]\), and then substitute any other value for \([value]\) (or equate coefficients of \([value]\)) to find \([value]\):

\([value]=\)

\([value]=\)

Finally express in partial fractions:

\(\dfrac{[value]}{([value])([value])}=\)
Feedback: The rational function is proper since the degree of the numerator is \(2\), and this is less than the degree of the denominator which is \(3\). The denominator is the product of a linear and (what we pretend to be) an irreducible quadratic. This lets use determine the antz and the problem is then to find constants \(A,\,B\) and \(C\) so that \[\dfrac{[value]}{([value])([value])}=\dfrac{[value]}{[value]}+\dfrac{[value]}{[value]}.\] We bring everything over a common denominator by multiplying both sides by \(([value])([value])\). This gives: \[[value]=[value].\] The easiest first step is to choose \([value]\) so that the linear factor \([value]=0\), i.e. we take \(\displaystyle{[value]=[value]}\). By substituting this in both sides of the equation we find \([value]=[value]\). Since there are no linear factors there are no real values we can substitute to eliminate all but one of the unknown constants. We can equate the coefficients of \([value]^2\) easily enough however, and we find \([value]=[value]\). We already have that \([value]=[value]\) and substitution gives \([value]=[value]\). If we substitute any other value for \([value]\), an easy option may be \([value]=0\) which is the same as equating the constant of the left and right side of the equation. In that case we find \([value]=[value]\). Using the values we found already this simplifies further to give \([value]=[value]\). We can piece everything together to see that \[\dfrac{[value]}{([value])([value])}=[value].\] I should note that since the quadratic is not irreducible we should go another step to write the term with the quadratic denominator in partial fractions. There's nothing wrong with the solution in this way, but one would never simplify in two stages. Assume the quadratic was irreducible and this is all fine.

Engineering Example

Admittance

Admittance, \(Y\), is a quantity which is used in analysing electronic circuits. A typical expression for admittance is \[ Y(s) = {s^2+4s+5\over (s^2+2s+4)(s+3)}\] where \(s\) can be thought of as representing frequency. To predict the behaviour of the circuit it is often necessary to express the admittance as the sum of its partial fractions and find the effect of each part separately. Express \(Y(s)\) in partial fractions.

The fraction is proper. The denominator contains an irreducible quadratic factor, which cannot be factorised further, and also a linear factor. Thus \[{s^2+4s+5\over (s^2+2s+4)(s+3)} = {As+B\over s^2+2s+4} + {C\over s+3} \qquad \qquad \qquad (1) \] Multiplying both sides of Equation (1) by \((s^2+2s+4)(s+3)\) we obtain \[ s^2+4s+5 = (As+B)(s+3) + C(s^2+2s+4) \qquad \qquad \qquad (2) \] To find the constant \(C\) we let \(s=-3\) in Equation (2) to eliminate \(A\) and \(B\). Thus \[ (-3)^2+4(-3)+5 = C((-3)^2+2(-3)+4)\] so that \[ 2 = 7C\qquad \text{and so}\qquad C={2\over 7}\] Equating coefficients of \(s^2\) in Equation (2) we find \( 1= A+C\) so that \(A=1-C= 1-\displaystyle{2\over 7} = \displaystyle{5\over 7}\). Equating constant terms in Equation (2) gives \( 5=3B+4C\) so that \[{3B=5-4C=5-4\left({2\over 7}\right)={27\over 7}}\] so \[ B=\displaystyle{9\over 7}\] Finally\[{Y(s) = {s^2+4s+5\over (s^2+2s+4)(s+3)} = {{5\over 7}s + {9\over 7} \over s^2+2s+4} + {{2\over 7}\over s+3}}\] which can be written as \[{Y(s) = {5s+9\over 7(s^2+2s+4)} + {2\over 7(s+3)}}\]