Advanced Higher Maths
Partial Fractions

This is a proper rational function, because the degree of the numerator (1) is less than the degree of the denominator (2). So we do not need to divide before starting.

The denominator contains two distinct (different) linear factors. So each partial fraction has a constant as its numerator. Let's call them A and B:

$$ \frac{3x+23}{(x-4)(x+3)} = \frac{A}{x-4} + \frac{B}{x+3} $$

Now multiply through by \((x-4)(x+3)\) to get rid of the algebraic fractions:

This identity is true for all values of \(x\) in the domain, so a convenient method of finding A and B is to substitute specific values of \(x\) that will eliminate one of the unknowns and let us find the other.

Let's start by seeing what \(x\!=\!4\) tells us about A:

$$ \begin{gather} 3(4)+23=A(4+3)+B(4-4)\\[6pt] 35=7A+0B\\[6pt] 35=7A\\[6pt] A=5 \end{gather} $$

Similarly, substituting \(x\!=\!-3\) will eliminate A and let us find B:

$$ \begin{gather} 3(-3)+23=A(-3+3)+B(-3-4)\\[6pt] 14=0A-7B\\[6pt] 14=-7B\\[6pt] B=-2\\[6pt] \end{gather} $$

Putting this together:

$$ \frac{3x+23}{(x-4)(x+3)} = \frac{5}{x-4} - \frac{2}{x+3} $$

First, we need to factorise the denominator:

Now we can see that the denominator contains two distinct linear factors. So each partial fraction has a constant as its numerator. Let's call them A and B:

$$ \frac{8x-21}{x(x-3)} = \frac{A}{x} + \frac{B}{x-3} $$

Now multiply through by \(x(x-3)\) to get rid of the fractions:

Now let's substitute specific values of \(x\) to eliminate each of the unknowns.

Substituting \(x\!=\!3\) will let us find B:

$$ \begin{gather} 8(3)-21=A(3-3)+3B\\[6pt] 3=0A+3B\\[6pt] 3=3B\\[6pt] B=1 \end{gather} $$

Similarly, substituting \(x\!=\!0\) will eliminate B and let us find A:

$$ \begin{gather} 8(0)-21=A(0-3)+0B\\[6pt] -21=-3A\\[6pt] A=7\\[6pt] \end{gather} $$

So the answer is:

$$ \frac{8x-21}{x^2-3x} = \frac{7}{x} + \frac{1}{x-3} $$

First, we need to factorise the denominator:

$$ \begin{eqnarray} \frac{2-4x}{x^3-3x^2+2x} &=& \frac{2-4x}{x(x^2-3x+2)} \\[6pt] &=& \frac{2-4x}{x(x-1)(x-2)} \end{eqnarray} $$

Again, the denominator contains distinct linear factors: three of them this time. So each partial fraction has a constant as its numerator. Let's call them A, B and C:

$$ \frac{2-4x}{x(x-1)(x-2)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x-2} $$

Multiply through by \(x(\!x-\!1)(x\!-\!2)\) to get rid of the fraction:

Substituting \(x\!=\!0\):

$$ \begin{gather} 2=A(-1)(-2)+0B+0C\\[6pt] 2=2A\\[6pt] A=1\\[6pt] \end{gather} $$

Substituting \(x\!=\!1\):

$$ \begin{gather} 2-4=0A+B(1)(-1)+0C\\[6pt] -2=-B\\[6pt] B=2\\[6pt] \end{gather} $$

Substituting \(x\!=\!2\):

$$ \begin{gather} 2-8=0A+0B=C(2)(1)\\[6pt] -6=2C\\[6pt] C=-3\\[6pt] \end{gather} $$

Putting this all together:

$$ \frac{2-4x}{x^3-3x^2+2x} = \frac{1}{x} + \frac{2}{x-1}-\frac{3}{x-2} $$

This denominator contains three linear factors, but they aren't distinct. The \(x^2\) is a repeated factor of \(x\small.\)

When this happens, we use the following form. Note how the repeated factor appears: once on its own and once squared.

$$ \frac{7x+2}{x^{2}(x+2)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+2} $$

Multiply through by \(x^{2}(x+2)\) to get rid of the fraction:

$$ 7x+2=Ax(x+2)+B(x+2)+Cx^2 $$

Substituting \(x\!=\!-2\) will let us find C:

$$ \begin{gather} -14+2=0+C(-2)^2\\[6pt] -12=4C\\[6pt] C=-3\\[6pt] \end{gather} $$

Next, \(x\!=\!0\) will let us find B.

$$ \begin{gather} 2=0+2B+0\\[6pt] B=1\\[6pt] \end{gather} $$

So far, we know the following:

We can find A by substituting any convenient value of \(x\). Let's just use \(x\!=\!1\small.\)

$$ \begin{gather} 7(1)+2=3A+1+2-3\\[6pt] 9=3A\\[6pt] A=3\\[6pt] \end{gather} $$

Note: Another way of finding A would have been to equate the \(x^2\) terms.

Either way, the partial fractions are:

$$ \frac{7x+2}{x^{2}(x+2)}=\frac{3}{x}+\frac{1}{x^2}-\frac{3}{x+2} $$

$$ \begin{eqnarray} \frac{1-5x}{x^3-2x^2+x} &=& \frac{1-5x}{x(x^2-2x+1)} \\[6pt] &=& \frac{1-5x}{x(x-1)^2} \end{eqnarray} $$

The denominator contains three linear factors, two of which are repeated. So:

$$ \begin{gather} \frac{1-5x}{x(x-1)^2}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{(x-1)^2}\\[6pt] 1-5x=A(x-1)^2+Bx(x-1)+Cx\\[6pt] \end{gather} $$

Substituting \(x\!=\!1\) will give us C:

$$ \begin{gather} 1-5=0+0+C\\[6pt] C=-4\\[6pt] \end{gather} $$

Substituting \(x\!=\!0\) will give us A:

$$ \begin{gather} 1=A(-1)^2+0+0\\[6pt] A=1\\[6pt] \end{gather} $$

To find B, we could make some other substitution, such as \(x\!=\!2,\) but it's probably simpler to use the method of equating coefficients. For example, equating the coefficients of \(x^2\) will give us B quite easily:

$$ \begin{gather} 0=A+B\\[6pt] 0=1+B\\[6pt] B=-1\\[6pt] \end{gather} $$

So the partial fractions are:

$$ \frac{1-5x}{x^3-2x^2+x} = \frac{1}{x} - \frac{1}{x-1} - \frac{4}{(x-1)^2} $$

We start by fully factorising the denominator:

This denominator contains an irreducible quadratic factor, so we need a linear expression as its numerator:

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

$$ 3x^2+2=A(x^2+2)+(Bx+C)x $$

Substituting \(x\!=\!0\) would let us find A, but there is no substitution that could find us B or C.

So instead of substituting, let's equate coefficients:

$$ \begin{eqnarray} \small\textsf{Equate}\ x^2\ \textsf{terms:}\normalsize\:\: && 3=A+B \\[6pt] \small\textsf{Equate}\ x\ \textsf{terms:}\normalsize\:\: && 0=C \\[6pt] \small\textsf{Equate constant terms:}\normalsize\:\: && 2=2A \\[6pt] && A=1 \end{eqnarray} $$

Substituting \(A\!=\!1\) into the \(x^2\) equation gives us \(B\!\!=2\small,\) and we're done.

So the final answer is:

$$ \frac{3x^2+2}{x^3+2x} = \frac{1}{x} + \frac{2x}{x^2+2} $$

This is our first example of an improper rational function, because the degree of the numerator is no longer lower than the degree of the denominator. So the first thing we have to do is to use polynomial long division:

Note that \(1\) is the quotient and \(-2x+8\) is the remainder. So:

Now we focus our attention on the proper rational function and express it as a sum of partial fractions:

$$ \begin{eqnarray} \frac{-2x+8}{x^2+x-2} &=& \frac{-2x+8}{(x+2)(x-1)} \\[6pt] &=& \frac{A}{x+2}+\frac{B}{x-1} \end{eqnarray} $$

Multiplying through to eliminate fractions:

Substituting \(x\!=\!1\):

$$ \begin{gather} -2+8=B(1+2)\\[6pt] 6=3B\\[6pt] B=2\\[6pt] \end{gather} $$

Substituting \(x\!=\!-2\):

$$ \begin{gather} 4+8=A(-2-1)\\[6pt] 12=-3A\\[6pt] A=-4\\[6pt] \end{gather} $$

So the final answer is:

$$ \frac{x^2-x+6}{x^2+x-2}\:=\:1 - \frac{4}{x+2}+\frac{2}{x-1} $$

This is a harder example of an improper rational function. You may be pleased to know that nothing quite as nasty has this has ever been set on an Advanced Higher Maths paper.

First we use algebraic long division to express the improper rational function as a polynomial plus a proper rational function.

So \(3x\) is the quotient and \(4x^2-2x+6\) is the remainder.

Now we need to factorise the denominator using Higher methods.

Check \(x\!=\!1\): \(\ 1^3-1^2+1-1=0\) so \(x\!-\!1\) is a factor and we can use synthetic division:

So \(x^3-x^2+x-1=(x-1)(x^2+1)\)

As \(x^2+1\) is an irreducible quadratic we are ready to set up our partial fractions:

Substituting \(x\!=\!1\):

$$ \begin{gather} 4-2+6=A(1^2+1)\\[6pt] 8=2A\\[6pt] A=4\\ \end{gather} $$

Equating the \(\raise 0.5pt{x^2}\) coefficients:

$$ \begin{gather} 4=A+B\\[6pt] 4=4+B\\[6pt] B=0\\ \end{gather} $$

Equating the \(\raise 0.5pt{x}\) coefficients:

$$ \begin{gather} -2=C-0\\[6pt] C=-2 \end{gather} $$

So the final answer is:

$$ 3x+\frac{4}{x-1}-\frac{2}{x^2+1} $$

The denominator contains an irreducible quadratic factor, so we need a linear expression as its numerator:

$$ \frac{3x^2-3x+5}{x(x^2+5)} = \frac{A}{x} + \frac{Bx+C}{x^2+5} $$

$$ 3x^2-3x+5=A(x^2+5)+(Bx+C)x $$

Substituting \(x=0\) lets us find A:

$$ \begin{gather} 5=5A+0\\[6pt] A=1\\ \end{gather} $$

Equating the coefficients of \(x^2\) gives us B:

$$ \begin{gather} 3=A+B\\[6pt] 3=1+B\\[6pt] B=2\\ \end{gather} $$

Equating the coefficients of \(x\) gives us C:

$$ \begin{gather} -3=C\\[6pt] C=-3 \end{gather} $$

So the final answer is:

$$ \frac{3x^2-3x+5}{x(x^2+5)} = \frac{1}{x} + \frac{2x-3}{x^2+5} $$

The denominator is already factorised and contains three linear factors, two of which are repeated. So:

$$ \frac{3x^2-x-14}{(x+3)(x-1)^2} = \frac{A}{x+3} + \frac{B}{x-1}+ \frac{C}{(x-1)^2} $$

$$ 3x^2-x-14=A(x-1)^2+B(x+3)(x-1)+C(x+3) $$

Substituting \(x\!=\!1\) will give us C:

Substituting \(x\!=\!-3\) will give us A:

$$ \begin{gather} 3(3^2)-(-3)-14=A(-3-1)^2+0+0\\[6pt] 27+3-14=16A\\[6pt] 16=16A\\[6pt] A=1\\[6pt] \end{gather} $$

To find B, we could make some other substitution, such as \(x\!=\!2,\) but it's probably simpler to use the method of equating coefficients. For example, equating the coefficients of \(x^2\) will give us B quite easily:

$$ \begin{gather} 3=A+B\\[6pt] 3=1+B\\[6pt] B=2\\[6pt] \end{gather} $$

So the partial fractions are:

$$ \frac{1}{x+3} + \frac{2}{x-1} - \frac{3}{(x-1)^2} $$

This was part (a) of a multi-part question which also involved integration by parts and a differential equation.

The denominator contains two distinct linear factors. So each partial fraction has a constant as its numerator. Let's call them A and B:

$$ \frac{-2}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1} $$

Now multiply through by \(x(x\!+\!1)\) to get rid of the algebraic fractions:

This identity is true for all values of \(x\) in the domain, so a convenient method of finding A and B is to substitute specific values of \(x\) that will eliminate one of the unknowns and let us find the other.

Let's start by substituting \(x\!=\!0\small.\) This will eliminate B:

$$ \begin{gather} -2=A(1)+B(0)\\[6pt] A=-2\\[6pt] \end{gather} $$

Similarly, substituting \(x\!=\!-1\) will eliminate A and let us find B:

$$ \begin{gather} -2=A(0)+B(-1)\\[6pt] B=2\\[6pt] \end{gather} $$

Putting this together:

$$ \frac{-2}{x(x+1)} = -\frac{2}{x} + \frac{2}{x+1} $$

The denominator is already factorised and contains three distinct linear factors.

So we start by expressing the given expression in the form:

$$ \frac{A}{x-1} + \frac{B}{x-3}+ \frac{C}{x+5} $$

Multiplying through by the denominator of the given expression, we obtain:

$$ \begin{eqnarray} 2x^2-18x+4 &=& A(x-3)(x+5)\\[6pt] &+& B(x-1)(x+5)\\[6pt] &+& C(x-1)(x-3)\\[6pt] \end{eqnarray} $$

Substituting carefully-chosen values of \(x\) will enable us to find each of the constants.

Substituting \(x\!=\!1\) will give us A:

$$ \begin{gather} 2-18+4=A(-2)(6)\\[6pt] -12=-12A\\[6pt] A=1\\ \end{gather} $$

Substituting \(x\!=\!3\) will give us B:

$$ \begin{gather} 18-54+4=B(2)(8)\\[6pt] -32=16A\\[6pt] B=-2\\ \end{gather} $$

Substituting \(x\!=\!-5\) will give us C:

$$ \begin{gather} 50+90+4=C(-6)(-8)\\[6pt] 144=48C\\[6pt] C=3\\ \end{gather} $$

So the partial fractions are:

$$ \frac{1}{x-1}-\frac{2}{x-3}+\frac{3}{x+5} $$