Partial Fraction Decomposition

 

Let $a, b, c, A, B, C \in \mathbb{R}$ and $m, n \in \mathbb{Z}^+$. If $ax^2 + bx + c$ is an irreducible polynomial, then rational expressions of the form

\[
\frac{A}{(ax + b)^m} \quad \text{and} \quad \frac{Bx + C}{(ax^2 + bx + c)^n}
\]

are called partial fractions.

 

Example:

 

\( \bullet \quad \) The following expressions are partial fractions:

\[ \frac{3}{(2x + 1)^4} , \quad \frac{2x + 1}{x^2 + x + 1} , \quad \frac{1}{x^2 + 1} \]

\( \bullet \quad \) The following expressions are not partial fractions:

\[ \frac{1}{x^2 – 4} , \quad \frac{2x}{(x – 1)(x – 3)} , \quad \frac{x^2}{x^2 + x + 1} \]

 

To express a rational function as a sum of partial fractions, we proceed with the following algebraic steps:

1) Factor the denominator of the rational expression into its irreducible components. These factors will form the denominators of the individual partial fractions.

2) Set up the numerator of each partial fraction as a polynomial whose degree is exactly one less than the degree of its respective denominator.

3) Find a common denominator to recombine the expressions, then equate the coefficients of the corresponding terms using the principle of polynomial equality to solve for the unknown constants.

 

Example:

 

Let us decompose the following rational expression into partial fractions:

\[
\frac{x + 2}{x^2 – 5x + 6}
\]

First, factor the denominator:

\[
\frac{x + 2}{(x – 2)(x – 3)} = \frac{A}{x – 2} + \frac{B}{x – 3}
\]

Combine the terms over a common denominator:

\[
= \frac{A (x – 3) + B (x – 2)}{(x – 2)(x – 3)}
\]

Since the denominators on both sides of the equation are identical, their numerators must also be equal. Thus,

\[
x + 2 = (A + B)\cdot x \;- \;3A \;- \; 2B
\]

By equating the coefficients, we obtain the system:

\[
A + B = 1 \quad \text{and} \quad -3A – 2B = 2
\]

Solving this system yields \( A = -4 \) and \( B = 5 \). Therefore,

\[
\frac{x + 2}{x^2 – 5x + 6} = \frac{-4}{x – 2} + \frac{5}{x – 3}
\]

 

Note:

 

If the roots of the polynomial in the denominator are distinct real numbers, the unknown constants can be determined evaluated directly by substituting the roots, bypassing the full method of equating coefficients (Heaviside cover-up method).

 

Example:

 

Let us decompose the following rational expression into partial fractions:

\[
\frac{x + 2}{x^3 – 3x^2 + 2x}
\]

Decompose the expression based on the linear factors of the denominator:

\[
\frac{x + 2}{x(x – 1)(x – 2)} = \frac{A}{x} + \frac{B}{x – 1} + \frac{C}{x – 2}
\]

 

Combine the fractions over a common denominator:

\[
\frac{x + 2}{x(x – 1)(x – 2)} = \frac{A(x – 1)(x – 2) + Bx(x – 2) + Cx(x – 1)}{x(x – 1)(x – 2)}
\]

 

Equating the numerators yields:

\[
\Rightarrow x + 2 = A(x – 1)(x – 2) + Bx(x – 2) + Cx(x – 1)
\]

Substituting the roots of the denominator, which are \( 0, 1, 2 \), into the equation above:

For \( x = 0 \):

\[
0 + 2 = A \cdot (0-1) \cdot (0-2) + 0 + 0 \Rightarrow A = 1
\]

For \( x = 1 \):

\[
1 + 2 = 0 + B \cdot 1 \cdot (1 – 2) + 0 \Rightarrow B = -3
\]

For \( x = 2 \):

\[
2 + 2 = 0 + 0 + C \cdot 2 \cdot (2 – 1) \Rightarrow C = 2
\]

Thus, the decomposition is:

\[
\frac{x + 2}{x^3 – 3x^2 + 2x} = \frac{1}{x} – \frac{3}{x – 1} + \frac{2}{x – 2}
\]

 

Example:

 

Let us decompose the following rational expression into partial fractions:

\[
\frac{1}{x^3 – x^2 + x – 1}
\]

Factoring the denominator by grouping gives:

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

Combine the terms over a common denominator:

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

Equating the numerators:

\[
\Rightarrow 1 = A (x^2 + 1) + (Bx + C) (x – 1)
\]

Substituting the real root \( x = 1 \) into the identity:

For \( x = 1 \):

\[
1 = A(1^2 + 1) + 0 \Rightarrow A = \frac{1}{2}
\]

Choosing \( x = 0 \) to evaluate the constants:

\[
1 = A(0^2 + 1) + (B \cdot 0 + C) \cdot (0 – 1) \Rightarrow C = -\frac{1}{2}
\]

Choosing \( x = 2 \):

\[
1 = \frac{1}{2} (2^2 + 1) + \left(B \cdot 2 – \frac{1}{2}\right) \cdot (2 – 1) \Rightarrow B = -\frac{1}{2}
\]

Substituting the calculated constants back into the decomposition:

\[
\frac{1}{x^3 – x^2 + x – 1} = \frac{\frac{1}{2} }{x – 1} + \frac{-\frac{1}{2}x\; -\; \frac{1}{2}}{x^2 + 1}
\]

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

 

QUESTION 18

 

Given the algebraic identity

\[
\frac{x^3}{x^2 \;- \; x \;-\; 2} = A x + B + \frac{C}{x + 1} + \frac{D}{x \;-\; 2}
\]

what is the value of the sum \( C + D \)?

\[
\text{A)} 1 \quad
\text{B) } 2 \quad
\text{C) } 3 \quad
\text{D) } 4 \quad
\text{E) } 5
\]

 

Solution:

 

Since the degree of the numerator is greater than or equal to the degree of the denominator, we perform polynomial long division:

\[
\begin{array}{r|l}
x^3 \phantom{aaaaaaaaa}& x^2 – x-2 \\
x^3\; – \;x^2 \;- \; 2x & \rule{35mm}{0.35mm} \\
– \rule{45mm}{0.35mm} & x+1 \\
x^2 + 2x \phantom{aa}&\\
x^2-x-2\\
– \rule{45mm}{0.35mm} \\
3x+2
\end{array}
\]

From the quotient and remainder, the expression can be rewritten as:

\[
\frac{x^3}{x^2 – x – 2} = x + 1 + \frac{3x + 2}{x^2 – x – 2}
\]

Matching this with the given identity structure gives $A=1$, $B=1$, and:

\[
= x + 1 + \frac{C}{x + 1} + \frac{D}{x – 2}
\]

Now, decompose the remaining proper rational fraction:

\[
\frac{3x + 2}{x^2 – x – 2} = \frac{C}{x + 1} + \frac{D}{x – 2}
\]

Solving for the coefficients gives:

\[
\Rightarrow C = \frac{1}{3} \quad \text{and} \quad D = \frac{8}{3}
\]

Evaluating the required sum:

\[
C + D = \frac{1}{3} + \frac{8}{3} = 3
\]

 

\(\textbf{Correct Answer: C} \)