UK A-Level: Algebra and Partial Fractions

Mastering algebraic manipulation, particularly the method of partial fractions, is a cornerstone of the A-Level mathematics curriculum. It transforms complex, intimidating expressions into simpler components that are far easier to analyze, integrate, or expand. This skill is the essential preprocessing step for integrating rational functions and finding binomial expansions, directly linking pure algebra to calculus and series.

Simplifying Algebraic Fractions: The Foundational Skill

Before you can decompose a fraction, you must be able to manipulate and simplify it. An algebraic fraction is simply a fraction where the numerator, denominator, or both contain algebraic expressions. The core principle is identical to numerical fractions: you can only add or subtract fractions with a common denominator, and you simplify by factorising and cancelling common factors that are common to the entire numerator and the entire denominator.

For example, simplify $\frac{x^2-9}{x^2+4x+3}$. The first step is always factorisation: $$\frac{x^2-9}{x^2+4x+3}=\frac{(x-3)(x+3)}{(x+3)(x+1)}$$ Here, $(x+3)$ is a common factor. Provided $x\ne -3$, you can cancel to obtain $\frac{x-3}{x+1}$. Crucially, you cannot cancel terms that are not factors. A common mistake is attempting to cancel the $x^2$ terms in the original form, which is invalid.

Polynomial Division and the Factor & Remainder Theorems

When the degree of the numerator is greater than or equal to the degree of the denominator, you have an improper algebraic fraction. Partial fractions require proper fractions, so your first step is often polynomial long division.

Polynomial division works analogously to numerical long division. For example, dividing $2x^3+3x^2-4x+1$ by $x^2+x-1$ yields a quotient of $2x+1$ and a remainder of $-3x+2$. This means: $$\frac{2x^3+3x^2-4x+1}{x^2+x-1}=(2x+1)+\frac{-3x+2}{x^2+x-1}$$ You then apply partial fractions to the proper fractional part, $\frac{-3x+2}{x^2+x-1}$.

The factor theorem states that if for a polynomial $f(x)$, $f(a)=0$, then $(x-a)$ is a factor of $f(x)$. This is indispensable for factorising the denominators you will encounter. Its close relative, the remainder theorem, states that when a polynomial $f(x)$ is divided by $(x-a)$, the remainder is $f(a)$. You can use this to quickly find remainders without performing full division, or to check potential factors.

Decomposition into Partial Fractions

The goal of partial fraction decomposition is to express a single complex fraction as a sum of simpler fractions. The form you use depends entirely on the factors in the denominator.

For linear factors that are all distinct, each factor $(px+q)$ in the denominator gives rise to a partial fraction of the form $\frac{A}{px+q}$. Consider: $$\frac{11x+12}{(x+1)(x+2)}\equiv \frac{A}{x+1}+\frac{B}{x+2}$$ To find $A$ and $B$, you multiply through by the common denominator $(x+1)(x+2)$: $$11x+12\equiv A(x+2)+B(x+1)$$ This is an identity, true for all values of $x$. You can solve by substituting strategic values: let $x=-1$ to get $11(-1)+12=A(1)\Rightarrow A=1$. Let $x=-2$ to get $11(-2)+12=B(-1)\Rightarrow B=10$. Thus: $$\frac{11x+12}{(x+1)(x+2)}\equiv \frac{1}{x+1}+\frac{10}{x+2}$$

For repeated linear factors, such as $(x-a{)}^2$, you must include terms for every power up to the repetition. For a factor $(x+1{)}^2$, the decomposition would include both $\frac{A}{x+1}$ and $\frac{B}{(x+1{)}^2}$.

A typical example with a repeated factor: $$\frac{5x^2+9x-4}{(x+1)(x-2{)}^2}\equiv \frac{A}{x+1}+\frac{B}{x-2}+\frac{C}{(x-2{)}^2}$$ You would multiply through by $(x+1)(x-2{)}^2$ and then use a combination of substitution (for $x=-1$ and $x=2$ to find $A$ and $C$ easily) and comparing coefficients to find $B$.

Application to Integration and Series

This algebraic process unlocks powerful techniques in calculus and algebra. In integration, a complex rational function like $\int \frac{11x+12}{(x+1)(x+2)}dx$ becomes straightforward when decomposed: $$\int \left(\frac{1}{x+1}+\frac{10}{x+2}\right)dx=\ln |x+1|+10\ln |x+2|+C$$ Without partial fractions, this integral would be very difficult to solve.

For binomial expansion and series work, you often need to expand expressions like $\frac{2x+5}{(1-x)(2+x)}$. First, decompose into partial fractions to get forms like $\frac{A}{1-x}$ and $\frac{B}{2+x}$, which are standard binomial forms $(1-X{)}^{-n}$ that you can expand easily using the binomial theorem for negative or fractional powers, valid within a specific radius of convergence.

Common Pitfalls

1. Forgetting to Check for Improper Fractions: Attempting partial fractions directly on an improper fraction will lead to an incorrect, unsolvable equation. Always perform polynomial long division first if the numerator's degree is greater than or equal to the denominator's.
2. Incorrect Form for Repeated Factors: For a denominator factor of $(x-3{)}^3$, you must use three terms: $\frac{A}{x-3}+\frac{B}{(x-3{)}^2}+\frac{C}{(x-3{)}^3}$. Using only $\frac{A}{(x-3{)}^3}$ is insufficient.
3. Algebraic Errors in Solving for Constants: When multiplying through by the denominator, ensure every term is correctly expanded. A reliable method is to use a mix of substitution (using zeros of the factors) and equating coefficients to minimize arithmetic mistakes.
4. Cancelling Before Decomposing: Avoid the temptation to cancel common factors across a sum of partial fractions. For instance, in $\frac{1}{x+1}+\frac{10}{x+2}$, you cannot cancel anything. Cancellation only applies to factors common to the entire numerator and entire denominator of a single fraction.

Summary

• Partial fraction decomposition is a critical algebraic technique for rewriting a complex rational expression as a sum of simpler fractions, categorised by the factors (linear or repeated) in the original denominator.
• The factor and remainder theorems are essential tools for factorising polynomials, which is the prerequisite first step for any decomposition.
• Always ensure the fraction is proper (numerator degree < denominator degree) by using polynomial long division first if necessary.
• The primary application in A-Level is to convert difficult integrals into a sum of simple logarithmic or power function integrals, and to prepare rational functions for binomial expansion.
• Mastery hinges on meticulous algebraic manipulation, correctly setting up the partial fraction form, and accurately solving for the unknown constants using substitution and coefficient comparison.