8.10 Partial Fractions

Example 1

Earlier, you were asked how to go from one fraction to multiple simpler fractions. To decompose the rational expression into the sum of two simpler fractions you need to use partial fraction decomposition.

\(\begin{aligned} \frac{4 x-9}{x^{2}-3 x} &=\frac{A}{x}+\frac{B}{x-3} \\ 4 x-9 &=A(x-3)+B x \\ 4 x-9 &=A x-3 A+B x \end{aligned}\)

Notice that the constant term -9 must be equal to the constant term \(-3 A\) and that the terms with \(x\) must be equal as well.

\(\begin{aligned}-9 &=-3 A \\ 4 &=A+B \end{aligned}\)

Solving this system yields:

\(A=3, \quad B=1\)

Therefore,

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

Example 2

Use partial fractions to decompose the following rational expression.

\(\frac{7 x^{2}+x+6}{x^{3}+3 x}\)

First factor the denominator and identify the denominators of the partial fractions.

\(\frac{7 x^{2}+x+6}{x\left(x^{2}+3\right)}=\frac{A}{x}+\frac{B x+C}{x^{2}+3}\)

When the fractions are eliminated by multiplying through by the LCD the equation becomes:

\(7 x^{2}+x+6=A\left(x^{2}+3\right)+x(B x+C)\)

\(7 x^{2}+x+6=A x^{2}+3 A+B x^{2}+C x\)

Notice the squared term, linear term and constant term form a system of three equations with three variables

\(A+B=7\)

\(C=1\)

\(3 A=6\)

In this case it is easy to see that \(A=2, B=5, C=1\). Often, the resulting system of equations is more complex and would benefit from your knowledge of solving systems using matrices.

\(\frac{7 x^{2}+x+6}{x\left(x^{2}+3\right)}=\frac{2}{x}+\frac{5 x+1}{x^{2}+3}\)

Example 3

Decompose the following rational expression.

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

First identify the denominators of the partial fractions.

\(\frac{5 x^{4}-3 x^{3}-x^{2}+4 x-1}{(x-1)^{3} x^{2}}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{(x-1)^{3}}+\frac{D}{x}+\frac{E}{x^{2}}\)

When the entire fraction is multiplied through by \((x-1)^{3} x^{2}\) the equation results to.

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

\(=A(x-1)^{2} x^{2}+B(x-1) x^{2}+C x^{2}+D(x-1)^{3} x+E(x-1)^{3}\)

Multiplication of each term can be done separately to be extra careful.

\(A x^{4}-2 A x^{3}+A x^{2}\)

\(B x^{3}-B x^{2}\)

\(C x^{2}\)

\(D x^{4}-3 D x^{3}+3 D x^{2}-D x\)

\(E x^{3}-3 E x^{2}+3 E x-E\)

Group terms with the same power of \(x\) and set equal to the corresponding term.

\(\begin{aligned} 5 x^{4} &=A x^{4}+D x^{4} \\ -3 x^{3} &=-2 A x^{3}+B x^{3}-3 D^{3}+E x^{3} \\ -x^{2} &=A x^{2}-B x^{2}+C x^{2}+3 D x^{2}-3 E x^{2} \\ 4 x &=-D x+3 E x \\ -1 &=-E \end{aligned}\)

From these 5 equations, every \(x\) can be divided out. Assume that \(x \neq 0\) because if it were, then the original expression would be undefined.

\(\begin{aligned} 5 &=A+D \\ -3 &=-2 A+B-3 D+E \\ -1 &=A-B+C+3 D-3 E \\ 4 &=-D+3 E \\ -1 &=E \end{aligned}\)

This is a system of equations of five variables and 5 equations. Some of the equations can be solved using logic and substitution like \(E=-1, D=-7, A=12\). You can use any method involving determinants or matrices. In this case it is easiest to substitute known values into equations with one unknown value to get more known values and repeat.

\(\begin{aligned} B &=1 \\ C &=6 \\ \frac{5 x^{4}-3 x^{3}-x^{2}+4 x-1}{(x-1)^{3} x^{2}} &=\frac{12}{x-1}+\frac{1}{(x-1)^{2}}+\frac{6}{(x-1)^{3}}+\frac{-7}{x}+\frac{-1}{x^{2}} \end{aligned}\)

Example 4

Decompose the following rational expression.

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

First identify the denominators of the partial fractions.

\(\frac{5 x^{4}-3 x^{3}-x^{2}+4 x-1}{(x-1)^{3} x^{2}}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{(x-1)^{3}}+\frac{D}{x}+\frac{E}{x^{2}}\)

When the entire fraction is multiplied through by \((x-1)^{3} x^{2}\) the equation results to.

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

\(=A(x-1)^{2} x^{2}+B(x-1) x^{2}+C x^{2}+D(x-1)^{3} x+E(x-1)^{3}\)

Multiplication of each term can be done separately to be extra careful.

\(A x^{4}-2 A x^{3}+A x^{2}\)

\(B x^{3}-B x^{2}\)

\(C x^{2}\)

\(D x^{4}-3 D x^{3}+3 D x^{2}-D x\)

\(E x^{3}-3 E x^{2}+3 E x-E\)

Group terms with the same power of \(x\) and set equal to the corresponding term.

\(\begin{aligned} 5 x^{4} &=A x^{4}+D x^{4} \\ -3 x^{3} &=-2 A x^{3}+B x^{3}-3 D^{3}+E x^{3} \\ -x^{2} &=A x^{2}-B x^{2}+C x^{2}+3 D x^{2}-3 E x^{2} \\ 4 x &=-D x+3 E x \\ -1 &=-E \end{aligned}\)

From these 5 equations, every \(x\) can be divided out. Assume that \(x \neq 0\) because if it were, then the original expression would be undefined.

\(\begin{aligned} 5 &=A+D \\ -3 &=-2 A+B-3 D+E \\ -1 &=A-B+C+3 D-3 E \\ 4 &=-D+3 E \\ -1 &=E \end{aligned}\)

This is a system of equations of five variables and 5 equations. Some of the equations can be solved using logic and substitution like \(E=-1, D=-7, A=12\). You can use any method involving determinants or matrices. In this case it is easiest to substitute known values into equations with one unknown value to get more known values and repeat.

\(\begin{aligned} B &=1 \\ C &=6 \\ \frac{5 x^{4}-3 x^{3}-x^{2}+4 x-1}{(x-1)^{3} x^{2}} &=\frac{12}{x-1}+\frac{1}{(x-1)^{2}}+\frac{6}{(x-1)^{3}}+\frac{-7}{x}+\frac{-1}{x^{2}} \end{aligned}\)

Example 5

Use matrices to help you decompose the following rational expression. Confirm the solution by adding the partial fractions.

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

\(\begin{aligned} \frac{5 x-2}{(2 x-1)(3 x+4)} &=\frac{A}{2 x-1}+\frac{B}{3 x+4} \\ 5 x-2 &=A(3 x+4)+B(2 x-1) \\ 5 x-2 &=3 A x+4 A+2 B x-B \\ 5 &=3 A+2 B \\-2 &=4 A-B \end{aligned}\)

\(\left[\begin{array}{cc|c}3 & 2 & 5 \\ 4 & -1 & -2\end{array}\right] \rightarrow \begin{array}{cc}\rightarrow & \cdot 4 & \rightarrow \\ \rightarrow & \cdot 3 & \rightarrow\end{array}\left[\begin{array}{cc|c}12 & 8 & 20 \\ 12 & -3 & -6\end{array}\right] \rightarrow \begin{array}{cc}\rightarrow \\ -I & \rightarrow\end{array}\left[\begin{array}{cc|c}12 & 8 & 20 \\ 0 & -11 & -26\end{array}\right] \rightarrow \begin{array}{cc}\rightarrow & 11 & \rightarrow \\ \rightarrow & .8 & \rightarrow\end{array}\left[\begin{array}{cc|c}132 & 88 & 220 \\ 0 & -88 & -208\end{array}\right]\)

\(\rightarrow I I \quad \rightarrow\)

\(\rightarrow\)\(\left[\begin{array}{cc|c}132 & 0 & 12 \\ 0 & -88 & -208\end{array}\right] \rightarrow \begin{array}{ccc}\rightarrow & \div 132 & \rightarrow \\ \rightarrow & \div-88 & \rightarrow\end{array}\left[\begin{array}{cc|c}1 & 0 & \frac{1}{11} \\ 0 & 1 & \frac{26}{11}\end{array}\right]\)

\(A=\frac{1}{11}, B=-\frac{26}{11}\)

\(\frac{5 x-2}{(2 x-1)(3 x+4)}=\frac{\frac{1}{11}}{2 x-1}+\frac{\frac{26}{11}}{3 x+4}\)

To confirm, add the fractions.

\(\begin{aligned} \frac{5 x-2}{(2 x-1)(3 x+4)} &=\frac{\frac{1}{11}}{2 x-1}+\frac{\frac{26}{11}}{3 x+4} \\ 5 x-2 &=\frac{1}{11}(3 x+4)+\frac{26}{11}(2 x-1) \\ 55 x-22 &=3 x+4+26(2 x-1) \\ 55 x-22 &=3 x+4+52 x-26 \\ 55 x-22 &=55 x-22 \end{aligned}\)