5. Partial fractions

After studying this section, you should be able to

• express an algebraic fraction in partial fractions.

5.1. Introduction

Example 5.1

Consider the following calculation.

\[\begin{align*} \frac {4}{3(x-1)} + \frac {5}{3(x+2)} &= \frac{4(x+2)}{3(x-1)(x+2)} + \frac{5(x-1)}{3(x-1)(x+2)}\\ &= \frac{4(x+2)+ 5(x-1)}{3(x-1)(x+2)}\\ &= \frac{9x +3}{3(x-1)(x+2)}\\ &= \frac{3x+1}{(x-1)(x+2)}. \end{align*}\]

Next semester we will meet calculations where it is much easier to work with the left hand side of the equation - two simple parts - than to work with the more complex right-hand side. The challenge in this section is, given a complicated fraction similar to the right-hand side, how do we split it up into the sum of two or more simpler things, like in the left-hand side?

We say that \(\displaystyle \frac{3x+1}{(x-1)(x+2)}\) is expressed in partial fractions as \(\displaystyle \frac {4}{3(x-1)} + \frac {5}{3(x+2)}\).

5.2. Examples

5.2.1. Non-repeated linear factors

Example 5.2

Express \(\frac{3x +1}{(x-1)(x+2)}\) in partial fractions.

Solution.
Note this is the same fraction as in Example 5.1 above. We use it to demonstrate the method.) It looks like we should be able to write this as the sum of a fraction with denominator \((x-1)\) and a fraction with denominator \((x+2)\). Let’s try to do this. Write

(5.1)\[\frac{3x +1}{(x-1)(x+2)} = \frac A{x-1} + \frac B{x+2}. \]

where \(A\) and \(B\) are unknown numbers to be found; that is, we try to find \(A\) and \(B\) such that the (5.1) holds for every value of \(x\). In fact, it can be shown that, as long as we follow certain methods which we will outline later, such numbers will exist.

We multiply (5.1) by \((x-1)(x+2),\) which is the denominator on the LHS, to get

\[label starstar 3x+1 = A(x+2) + B(x-1).\]

We now need to find \(A\) and \(B\) from this equation. To do this we pick ‘nice’ values for \(x\).

Put \(x = 1\) to get \(3+1=A(1+2)+B(1-1)\) so that \(4 = 3A + 0\) and hence \(A=\frac 43\).

Put \(x = -2\) to get \(3(-2)+1=A(-2+2)+B(-2-1)\) so that \(-5= 0 -3B\) and hence \(B =\frac 53\).

(Note: we chose \(x\) so that, in each case, one of the summands on the RHS of (5.1) vanished.)

Hence

\[ \frac{3x+1}{(x-1)(x+2)} = \frac{\frac{4}3}{x-1} + \frac{\frac53}{x+2} = \frac{4}{3(x-1)} + \frac{5}{3(x+2)}. \]

We now check our answer. For this we have two possibilities:

• We put another value for \(x\) in, say \(x = 0\) (one that is different from any we have already used). Then the LHS is equal to

\[ \frac{3\times 0 + 1}{(0-1)(0+2)} = \frac{1}{-2} = -\frac 12. \]

On the other hand, the RHS gives

\[ \frac{4}{3(0-1)} + \frac{5}{3(0+2)} = -\frac{4}{3} + \frac{5}{6} = -\frac 86 + \frac 56 = -\frac 36 = -\frac 12, \]

which is equal to the LHS so everything looks OK.

• We could start with the RHS and and add them together, putting the two terms on a common denominator, and check that we end up with the LHS.

Note

The first method of checking is usually quickest and is fine for exams.

Rule: The number of variables (letters), \(A,\) \(B,\dots\) etc, is always equal to the degree of the denominator in the starting fraction. This is a useful thing to remember if things seem to be going wrong with your calculation.

Exercise

Express the following in partial fractions.

(i) \(\displaystyle \frac{3x+1}{(x+2)(x+1)(x-3)}\);

(ii) \(\displaystyle \frac{x-1}{3x^2 -11x +10}\);

(iii) \(\displaystyle \frac{x}{(2+x)(2-x)}\);

(iv) \(\displaystyle \frac{6}{(x+3)(x-3)}\);

(v) \(\displaystyle \frac{3-4x}{2+3x-2x^2}\).

(Hint: Factorise (ii) by completing the square.)

Solution

(i) \(\frac{3x+1}{(x+2)(x+1)(x-3)}= \displaystyle\frac{1}{2(x+1)} - \frac 1{x+2}+ \frac 1{2(x-3)}\),

(ii) \(\displaystyle \frac{x-1}{3x^2 -11x +10}=\displaystyle \frac 1{x-2} - \frac 2{3x-5}\),

(iii) \(\displaystyle \frac{x}{(2+x)(2-x)}=\displaystyle \frac1{2(2-x)} - \frac 1{2(2+x)}\),

(iv) \(\displaystyle \frac{6}{(x+3)(x-3)}=\displaystyle \frac 1{x-3} - \frac 1{x+3}\),

(v) \(\displaystyle \frac{3-4x}{2+3x-2x^2}=\displaystyle \frac 2{1+2x} - \frac 1{2-x}\).

(Notes: for (ii), \(3x^2 -11 x +10 = 3(x^2 - \frac{11}{3} x) + 10 = 3((x- \frac{11}{6})^2- \frac{121}{36}) + \frac{120}{12} = 3(x- \frac {11}{6})^2 - \frac 1{12}\). Thus \(3x^2 - 11x + 10 = 0\) if and only if \(x- \frac{11}{6} = \pm \sqrt{\frac 1{36}} = \pm \frac 16\). Hence \(x_{1/2} = \frac {11}{6} \pm \frac 16\). Thus \(x = 2\) or \(x = \frac {10}{6} = \frac 53\).

For (v), \(-2x^2+3x+2 = -(2x^2-3x-2) = -(2x+1)(x-2) = (2x+1)(2-x)\). )

5.2.2. Quadratic factors that do not factorise

The method in the previous section will work whenever the denominator is a product of distinct linear factors (and the numerator has degree less than that of the denominator). We now see how to deal with factors in the denominator which don’t factorise over the real numbers.

Example 5.3

Express \(\frac{1}{(x^2+1)(x-1)}\) in partial fractions.

Solution.
Note that \(x^2 +1\) does not factorise over the real numbers. We write

(5.2)\[\frac{1}{(x^2+1)(x-1)} = \frac{Ax+B}{x^2+1} +\frac C{x-1}\]

where \(A,\) \(B,\) and \(C\) are real numbers to be found. Note the numerator of \(Ax+B\) over the factor that does not factorise.

Multiply (5.2) by the denominator \((x^2 +1)(x-1)\) to get

(5.3)\[1 = (Ax+B)(x-1) +C(x^2+1).\]

Put \(x =1\) to get \(1 = C(1+1) = 2C\). Hence \(C = \frac 12\).

Note that now there is no other nice value of \(x\) to put in. Hence we compare the coefficients of the LHS and the RHS of (5.3).

The coefficient of \(x^2\) on the LHS is \(0\) and on the RHS is \(A+C\). Hence \(0 = A+C\). As \(C = \frac 12,\) we find \(A = -\frac 12\).

The constant term on the LHS is \(1\) and on the RHS is \(-B+C\). So \(1 = -B + C\). As \(C = \frac 12,\) we get \(B = -\frac 12\).

It follows that \(A = -\frac 12,~B= -\frac 12\text{and }C=\frac 12\). Thus we get

\[\begin{align*} \frac 1{(x^2+1)(x-1)} &= \frac{-\frac 12 x - \frac 12}{x^2+1} + \frac{\frac 12}{x-1}~ = ~ - \frac {x +1}{2(x^2+1)} + \frac{1}{2(x-1)}. \end{align*}\]

We check our result. Note that we cannot choose \(x = 0\) as this corresponds to comparing the constant terms, which we have already used in our calculation. Putting in \(x = -1\) gives

\[ \mbox{LHS} = \frac{1}{(2)\times(-2)} = -\frac 14, \]

and

\[ \mbox{RHS} = -\frac{0}{2\times 2} + \frac 1{2\times (-2)} = - \frac 14 = \mbox{LHS}, \]

as hoped.

Exercise

Express \(\displaystyle \frac{x^2 +x+4}{x^3 +4x^2 +7x +6}\) in partial fractions.

Solution

We first have to factorise \(x^3 +4x^2 +7x +6\). Using the Factor Theorem, we see that \((x+2)\) is a factor and using long division (Section 3.3) we find that \(x^3 +4x^2 +7x + 6 = (x+2)(x^2 +2x +3)\). We next try to factorise \(x^2 +2x +3\). In fact, completing the square gives \(x^2 +2x +3 = (x+1)^2 -1 +3 = (x+1)^2 +2\). So this polynomial does not factorise. (Alternatively, look at the discriminant, \(\Delta\).)

We write

(5.4)\[\frac{x^2 +x+4}{(x+2)(x^2+2x+3)} = \frac{Ax+B}{x^2+2x+3} + \frac{C}{x+2}.\]

Multiply (5.4) by the denominator of the LHS to get

\[ x^2 +x +4 = (Ax +B)(x+2) + C(x^2 + 2x+3). \]

Put \(x = -2\) to get

\[\begin{split} \begin{array}{crcl} &(-2)^2 + (-2) + 4 &=& 0 + C((-2)^2 + 2 \times(-2) + 3)\\ \end{array} \end{split}\]

so that \(6=3C\) and hence \(C=2\).

Now compare coefficients. The coefficient of \(x^2\) on the LHS is \(1\) and on the RHS is \(A+C\). So \(1 = A+C\). As \(C = 2\) we find that \(A = -1\).

The constant term on the LHS is \(4\) and on the RHS is \(2B+3C\). So \(4 = 2B + 3C\). As \(C = 2\) we get \(B = -1\).

Hence \(A = -1\), \(B=-1\) and \(C= 2\). Thus we get

\[\begin{align*} \frac{x^2 +x+4}{x^3 +4x^2 +7x +6} &= \frac{-x -1}{x^2 +2x +3} + \frac 2{x+2}~=~-\frac{x +1}{x^2 +2x +3} + \frac 2{x+2}. \end{align*}\]

We check our result by putting in \(x = 1\) to get \(\mbox{LHS} = \frac{6}{18} = \frac 13,\) and \(\mbox{RHS} = -\frac{2}{6} + \frac{2}{3} = \frac 13 = \mbox{LHS}\). (Note that as we had compared constant terms above, again we had to avoid the value \(x=0\) in the check.)

Exercise

Express the following in partial fractions:

(i) \(\displaystyle \frac{6-x}{(1-x)(4+x^2)}\);

(ii) \(\displaystyle \frac{4}{(x+1)(2x^2+x+3)}\);

(iii) \(\displaystyle \frac{5x+2}{(x+1)(x^2-4)}\);

(iv) \(\displaystyle \frac{3+2x}{(2-x)(3+x^2)}\).

Solution

(i) \(\displaystyle \frac{6-x}{(1-x)(4+x^2)}=\displaystyle \frac{1}{1-x} + \frac {2+x}{4+x^2}\).

(ii) \(\displaystyle \frac{4}{(x+1)(2x^2+x+3)}=\displaystyle \frac{1}{x+1} - \frac{2x-1}{2x^2 +x +3}\).

(iii) \(\displaystyle \frac{5x+2}{(x+1)(x^2-4)}&=&\displaystyle \frac{1}{x+1} + \frac 1{x-2} - \frac2{x+2}\).

(iv) \(\displaystyle \frac{3+2x}{(2-x)(3+x^2)}=\displaystyle \frac 1{2-x} + \frac x{3+x^2}\).

5.2.3. Repeated linear factors

We now look at the case where we have repeated linear factors occurring.

Example 5.4

Express \(\frac{-x^2 -8x +25}{(x+1)(x^2-6x+9)}\) in partial fractions.

Solution.
Note first that \(x^2-6x+9=(x-3)^2\) using the binomial formula, so that we have a repeated factor of \((x-3)\) in the denominator. Remembering the rule from earlier that the numbers of variables we introduce must be equal to the degree of the denominator, we write

(5.5)\[\frac{-x^2 -8x +25}{(x+1)(x-3)^2} = \frac A{x+1} + \frac B{x-3} +\frac C{(x-3)^2}. \]

Multiply (5.5) by the denominator of the LHS to get

(5.6)\[-x^2 -8x +25 = A(x-3)^2 + B(x+1)(x-3) + C(x+1).\]

We try to find \(A,\) \(B,\) and \(C\). Put \(x = 3\) to get \(-9-24+25 = 4C\) so that \(-8 = 4C\) and so \(C=-2\).

Put \(x = -1\) to get \(-1 +8 +25 = A(-4)^2\) so that \(32 = 16A\) and \(A = 2\).

We now compare the coefficients of \(x^2\) of (5.6). The coefficient of \(x^2\) on the LHS is \(-1\) and on the RHS is \(A+B,\) so \(-1 = A+B\). As \(A = 2\) we get \(B = -3\).

Hence \(A = 2\), \(B=-3\) and \(C=-2\). Thus we get

\[\begin{align*} \frac{-x^2 -8x +25}{(x+1)(x-3)^2} = \frac 2{x+1} - \frac 3{x-3} -\frac 2{(x-3)^2}. \end{align*}\]

We check our result by putting in \(x = 0\) to get \(\mbox{LHS} = \frac{25}{9}\) and

\[ \mbox{RHS} = \frac{2}{1} + \frac 3{3} -\frac 29 = 3 -\frac29 = \frac{25}9 = \mbox{LHS}. \]

(Note here that \(x=0\) is a valid choice of \(x\) as we didn’t compare the constant terms in the process of the calculation.)

Exercise

Express \(\frac{3x^4 +9x^3 -x^2 -10x -1}{(x-1)^2 (x+2)^3}\) in partial fractions. (The full exercise is tricky and harder than you’ll get on any exam! As a simpler exercise, stop after writing the first equation (5.7). )

Solution

We write

(5.7)\[\frac{3x^4 +9x^3 -x^2 -10x -1}{(x-1)^2 (x+2)^3} = \frac{A}{x-1} + \frac B{(x-1)^2} + \frac C{x+2} + \frac D{(x+2)^2}+ \frac E{(x+2)^3}.\]

Multiply (5.7) by the denominator of the left-hand side to get

(5.8)\[\begin{split}\begin{array}{rcl} 3x^4 +9x^3 - x^2 -10x -1 &=& A(x-1)(x+2)^3 \\ &&+B(x+2)^3 + C(x-1)^2(x+2)^2 \\ &&+ D(x-1)^2(x+2)+E(x-1)^2\end{array}\end{split}\]

We try to find \(A,\) \(B,\) \(C,\) \(D\) and \(E\).

Put \(x = 1\) to get

\[\begin{split} \begin{array}{crcl} &3+9-1-10-1 &=& B(3)^3\\ \therefore&0 &=& 27B\\ \therefore&B &=& 0. \end{array} \end{split}\]

Put \(x = -2\) to get

\[\begin{split} \begin{array}{crcl} &3\times 16 - 9\times8 - 4+20-1 &=& E(-3)^2\\ \therefore&48 -72 +20 -5&=& 9E\\ \therefore&-9&=&9E\\ \therefore&E &=& -1. \end{array} \end{split}\]

We can easily compare the coefficients of \(x^4\): on the LHS, the coefficient is \(3\) and, on the RHS, it is \(A+C\). Hence \(3 = A+C\).

To compare the other coefficients would mean multiplying out the RHS of (5.8), which could be tiresome. Instead, we choose some more numbers to put in. Note that we already know that \(B = 0\) and \(E= -1\).

Put \(x = 0\) to get

\[\begin{split} \begin{array}{crcl} &-1 &=& A(-1)(2)^3 + C(-1)^2(2)^2 + D(-1)^2(2) -(-1)^2\\ \therefore&-1&=& -8A + 4C +2D -1\\ \therefore&0&=& -8A +4C +2D\\ \therefore&0&=& - 4A + 2C + D. \end{array} \end{split}\]

Put \(x = -1\) to get

\[\begin{split} \begin{array}{crcl} & 3- 9 -1 +10 -1&=& A(-2)(1)^3 + C(-2)^2(1)^2 + D(-2)^2(1) -(-2)^2\\ \therefore&2&=& -2A + 4C +4D -4\\ \therefore&6&=& -2A +4C +4D\\ \therefore&3&=& -A + 2C + 2D. \end{array} \end{split}\]

Hence we have three equations, namely

\[\begin{split} \begin{array}{cccccll} A &+ C& &=& 3 & ~\qquad\qquad & (1)\\ -4A &+2C &+D &=& 0 & & (2)\\ -A &+2C &+2D &=& 3 & & (3). \end{array} \end{split}\]

Multiply the equation (1) by \(4\) and add this to (2) to get an new equation (labelled (4)). Also add equation (1) to equation (3) to get a new equation, (5).

\[\begin{split} \begin{array}{cccccll} A &+ C& &=& 3 & ~\qquad\qquad & (1)\\ & 6C &+D &=& 12 & & (4)\\ &3C &+2D &=& 6 & & (5). \end{array} \end{split}\]

Next multiply (5) by \(-2\) and add this to (4) to get the new equation, (6).

\[\begin{split} \begin{array}{cccccll} A &+ C& &=& 3 & ~\qquad\qquad & (1)\\ & 6C &+D &=& 12 & & (4)\\ & &-3D &=& 0 & & (6). \end{array} \end{split}\]

Hence, from (6), \(D = 0\). Putting this into the (4) gives \(6C = 12\) so that \(C = 2\). Now equation (1) gives \(A = 1\). Hence the solutions are

\[ A = 1,~B = 0, ~ C = 2,~ D = 0\text{ and } E = -1. \]

Thus we finally get

\[ \frac{3x^4 +9x^3 -x^2 -10x -1}{(x-1)^2 (x+2)^3} = \frac{1}{x-1} + \frac 2{x+2} - \frac 1{(x+2)^3}. \]

Note: It is a good exercise for you to check that the result is true by putting the RHS over a common denominator \((x-1)^2(x+2)^3\).

Example 5.5

Express \(\frac{x^6+3x^5+3x^4+3x^3+3x^2+2x+2}{(x^2+2)x^5}\) in partial fractions.

Solution.

Note that, as with all our example so far, the fraction is bottom heavy: the degree of the numerator is 6 and the degree of the denominator is 7. Further, the quadratic \(x^2+2\) does not factorise further. Hence we write

\[ \frac{x^6+3x^5+3x^4+3x^3+3x^2+2x+2}{(x^2+2)x^5} = \frac{Ax + B}{x^2+2} + \frac Cx + \frac D{x^2} + \frac E{x^3} + \frac{F}{x^4} + \frac G{x^5} \]

and multiplying through by the denominator of the LHS gives

\[\begin{align*} x^6+3x^5+3x^4+3x^3+3x^2+2x+2 & = & (Ax+B)x^5 + C(x^2+2)x^4 + D(x^2+2)x^3\\ & & + E(x^2+2)x^2+ F(x^2+2)x+ G(x^2+2). \end{align*}\]

Putting \(x = 0\) gives \(2 = 2G\) so that \(G = 1\). There are no other very easy values of \(x\) to choose, so we equate coefficients.

• From \(x^6\) we get \(1 = A+C\).

• From \(x^5\) we get \(3 = B +D\).

• From \(x^4\) we get \(3 = 2C+E\).

• From \(x^3\) we get \(3 = 2D+F\).

• From \(x^2\) we get \(3 = 2E+G = 2E +1\) so that \(E = 1\).

• From \(x\) we get \(2= 2F\) so that \(F = 1\).

Hence \(D = 1\) (from the \(x^3\) coefficient) and so \(B=2\) (from \(x^5\)).

The \(x^4\) coefficient together with \(E=1\) gives \(C = 1\).

Finally, from the \(x^6\) coefficient, \(A= 0\).

Hence

\[ \frac{x^6+3x^5+3x^4+3x^3+3x^2+2x+2}{(x^2+2)x^5} = \frac{2}{x^2+2} + \frac 1x + \frac 1{x^2} + \frac 1{x^3} + \frac{1}{x^4} + \frac 1{x^5}. \]

We check our answer by putting in \(x =1\) into the equation to get

\[ \mbox{LHS} = \frac{1+3+3+3+3+2+2}{3} = \frac{17}3 \]

and

\[ \mbox{RHS} = \frac{2}{3} +1+1+1+1+1 = \frac{2}{3} + \frac{15 }{3} = \frac{17}{3} = \mbox{LHS} \]

Exercise

1. Express \(\displaystyle \frac{x+1}{(x+3)^2}\) and \(\displaystyle \frac{2x^2 -5x +7}{(x-2)(x-1)^2}\) in partial fractions.

2. Find the values of \(A,\) \(B,\) \(C\) and \(D,\) if

\[ \frac {x^3 -10x^2 +26 x+3}{(x+3)(x-1)^3} = \frac A{x+3} + \frac B{x-1} + \frac C{(x-1)^2} + \frac D{(x-1)^3}. \]

5.3. Improper fractions

So far we have dealt with fractions that were bottom heavy: the degree of the numerator was less than the degree of the denominator. This may not always be the case. We say that an algebraic fraction is improper if the degree of the numerator is greater than or equal to the degree of the denominator.

Example 5.6

Express \(\frac{2x^2+x+1}{(x-1)(x+1)}\) in partial fractions.

Solution.
Note that the fraction is improper: the degree of the numerator is 2, which is equal to the degree of the denominator. When this occurs we must use long division first or our earlier methods will not work. Here is the method.

Step 1 Multiply out the denominator (if necessary) to get \((x-1)(x+1) = x^2 -1\).

Step 2: Divide the numerator by the denominator using long division.

\[\begin{split} \require{enclose} \begin{array}{rl} 2 &\\[3pt] x^2+ 0x -1\enclose{longdiv}{ 2x^2+x+1}\kern-.2ex& \\ \underline{-(2x^2+0x-2) } &\\[3pt] x+3 \end{array} \end{split}\]

Thus \(2x^2 +x +1 = 2(x^2-1)+ (x+3)\).

Step 3: We can now simplify to get

\[ \frac{2x^2 +x +1}{(x-1)(x+1)} = \frac{2(x-1)(x+1) + (x+3)}{(x-1)(x+1)}= 2 +\frac{x+3}{(x-1)(x+1)}. \]

Step 4: Express \(\frac{x+3}{(x-1)(x+1)}\) in partial fractions using our normal methods. We write

(5.10)\[\frac{x+3}{(x-1)(x+1)} = \frac A{x-1} + \frac B{x+1}. \]

Multiply by the denominator to get

(5.10)\[x+3 = A(x+1) + B(x-1).\]

Put \(x = 1\) into (5.10) to get \(1+3 = A(1+1)\) so that \(4 = 2A\) and hence \(A=2\).

Put \(x = -1\) into (5.10) to get \(-1+3 = B(-1-1)\) so that \(2 = -2B\) and hence \(B = -1\).

Thus \(A= 2\) and \(B = -1\) and we get

\[ \frac{x+3}{(x-1)(x+1)} = \frac 2{x-1} - \frac {1}{x+1}. \]

Hence, returning to our original fraction, we have

\[ \frac{2x^2+x+1}{(x-1)(x+1)} = 2 + \frac 2{x-1} - \frac{1}{x+1} \]

is the expression written in partial fractions, and we are done.

We check our answer by putting in \(x = 0\) to get \(\mbox{LHS} = \frac{1}{(-1)(1)} = -1\) and \(\mbox{RHS} = 2 + \frac 2{-1} - \frac 1{1} = -1 = \mbox{LHS}\).

Exercise

Express \(\frac {x^5 +x}{(x+1)(x^2+x+1)}\) as partial fractions.

Solution

Note that the fraction is improper.

Step 1: We multiply out the denominator to get

\[ (x+1)(x^2 + x+1) = x^3 +x^2 + x + x^2 +x + 1= x^3 +2x^2 +2x +1. \]

Step 2: Divide the numerator by the denominator in a long division.

\[\begin{split} \require{enclose} \begin{array}{rl} x^2 -2x+2 & \\[3pt] x^3 +2x^2 +2x +1\enclose{longdiv}{x^5+0x^4+0x^3+0x^2+x+0}\kern-.2ex&\\ \underline{-(x^5+2x^4+2x^3+x^2) \hspace{4.5em}} &\\[3pt] -2x^4-2x^3-x^2+x & +0\\ \underline{-(-2x^4-4x^3-4x^2-2x) \hspace{1.5em}} &\\[3pt] 2x^3+3x^2+3x+0\\ \underline{-(2x^3+4x^2+4x+2)} &\\[3pt] -x^2-x-2 \end{array} \end{split}\]

Thus

\[\begin{align*} x^5 +x & = & (x^3 +2x^2 +2x +1)(x^2 -2x +2) - x^2 -x -2\\ &= & (x+1)(x^2+x+1)(x^2 -2x +2) -(x^2 +x +2). \end{align*}\]

Step 3: Simplify the original fraction to get

\[\begin{align*} \frac {x^5 +x}{(x+1)(x^2+x+1)} &= \frac{(x+1)(x^2+x+1)(x^2 -2x +2) -(x^2 +x +2)}{(x+1)(x^2+x+1)}\\ &= x^2 -2x +2 - \frac{x^2 +x +2}{(x+1)(x^2+x+1)}. \end{align*}\]

Step 4: Express \(\displaystyle \frac{x^2 +x +2}{(x+1)(x^2+x+1)}\) in partial fractions.

Firstly, we check if \(x^2 +x +1\) can be factorised. We have \(\Delta=1^2-4\times 1\times 1=-2<0\), so \(x^2 +x +1\) cannot be factorised over real numbers. We write

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

Multiply by the denominator of the LHS to get

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

Put \(x = -1\) into (5.11) to get \(1-1+2=A(1-1+1)\) so that \(2 = A\).

Next we compare coefficients. The coefficient of \(x^2\) on the LHS is \(1\) and on the RHS is \(A +B,\) so \(1 = A+B\). As \(A = 2,\) we get \(B = -1\).

The constant term on the LHS is \(2\) and on the RHS is \(A +C,\) so \(2 = A + C\). As \(A = 2\) we find that \(C = 0\).

Hence

\[ A = 2,~B= -1\text{ and }C= 0. \]

Thus

\[ \frac{x^2+x+2}{(x+1)(x^2+x+1)} = \frac 2{x+1} - \frac {x}{x^2+x+1}. \]

So the final answer is

\[\begin{align*} \frac {x^5 +x}{(x+1)(x^2+x+1)} &= x^2 -2x +2 - \frac{x^2 +x +2}{(x+1)(x^2+x+1)}\\ &= x^2 -2x +2 - \left( \frac 2{x+1} - \frac {x}{x^2+x+1}\right)\\ &= x^2 -2x +2 - \frac 2{x+1} + \frac {x}{x^2+x+1}. \end{align*}\]

We check our answer by putting in \(x = 1\) to get \(\mbox{LHS} = \frac{2}{(2)(3)} = \frac 13\) and \(\mbox{RHS} = 1 -\frac{2}{2} + \frac{1}{3} = \frac 13 = \mbox{LHS}\).

5.4. Recipe for Creating Partial Fractions

5.4.1. Workflow

Write \(\frac{f(x)}{g(x)}\) in partial fractions (where \(f(x)\) and \(g(x)\) are polynomials).

Step 1: If the degree of \(f(x)\) is greater than or equal to the degree of \(g(x)\) then we use long division to divide \(f(x)\) by \(g(x)\) to obtain

\[ \frac{f(x)}{g(x)} = q(x) + \frac{r(x)}{g(x)} \]

where \(q(x)\) and \(r(x)\) are polynomials and the degree of \(r(x)\) is less than the degree of \(g(x)\).

Step 2: Factorise \(g(x)\) into linear factors and quadratic factors that cannot be factorised any further. (NB: to test whether a quadratic can be factorised, look at the discriminant \(\Delta\).)

Step 3: For each non-repeated linear factor \((ax+b)\) of the denominator we use a partial fraction of the form \(\frac A{ax+b},\) where \(A\) is an unknown number we have to find.

Step 4: For each repeated linear factor \((ax+b)^r\) we use partial fractions of the form \(\frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + \dots + \frac{A_r}{(ax+b)^r},\) where \(A_1,\) \(A_2, \dots, A_r\) are unknown numbers we have to find.

Step 5: For each non-repeated quadratic factor \(ax^2 +bx+c\) that cannot be factorised, we use a partial fraction of the form \(\frac{Bx +C}{ax^2 +bx+c},\) where \(B\) and \(C\) are unknown numbers we have to find.

Step 6: We get the expression

(5.12)\[\frac{r(x)}{g(x)} = \mbox{partial fraction}(1) + \mbox{partial fraction}(2)+ \dots + \mbox{partial fraction}(n). \]

Step 7: Multiply (5.12) by the denominator of the LHS, that is by \(g(x)\).

Step 8: Find the unknown numbers \(A,\) \(B,\) \(C,\dots\) by a combination of choosing particular values for \(x\) and comparing coefficients.

Step 9: When you have all the values for the unknown numbers \(A,\) \(B,\) \(C,\dots\), put them into the equation (5.12) of Step6 to get \(\frac{r(x)}{g(x)}\) written as partial fractions.

Step 10: Use your result from Step~9 to get the final answer \(\frac{f(x)}{g(x)} = q(x) + \frac{r(x)}{g(x)}\) in partial fractions.

Step 11: Check your answer, either by putting in a value of \(x\) not used earlier and checking that both sides give the same or by combining the RHS of you final answer into a single fraction. That single fraction should turn out to be the LHS of your answer.