Partial fractionsAQA A-Level Maths: Revision notes

Adding fractions gives one fraction over a common denominator. Partial fractions reverse this: a single rational function is written as a sum of simpler fractions, each with one linear factor of the denominator. In this course the denominator is a product of up to three linear factors, with a factor allowed to be squared, and the numerator is a constant or a linear expression. The forms are:

• distinct factors: $\dfrac{px+q}{(x+a)(x+b)}\equiv\dfrac{A}{x+a}+\dfrac{B}{x+b}$
• a repeated factor: $\dfrac{px+q}{(x+a)^2}\equiv\dfrac{P}{x+a}+\dfrac{Q}{(x+a)^2}$

Write the form, multiply through by the denominator to get an identity, then substitute the value of $x$ that makes each factor zero. Example: $\dfrac{7x+1}{(x-1)(x+2)}\equiv\dfrac{A}{x-1}+\dfrac{B}{x+2}$ gives $7x+1\equiv A(x+2)+B(x-1)$. $x=1$: $8=3A$, so $A=\frac83$. $x=-2$: $-13=-3B$, so $B=\frac{13}{3}$.

With three distinct factors, use three substitutions. For $\dfrac{30}{(x+1)(x-2)(x+3)}$ write $30\equiv A(x-2)(x+3)+B(x+1)(x+3)+C(x+1)(x-2)$. $x=-1$: $30=-6A$, $A=-5$. $x=2$: $30=15B$, $B=2$. $x=-3$: $30=10C$, $C=3$. So the fraction equals $-\dfrac{5}{x+1}+\dfrac{2}{x-2}+\dfrac{3}{x+3}$.

A squared factor $(x+a)^2$ needs two terms: $\dfrac{B}{x+a}+\dfrac{C}{(x+a)^2}$. Leaving out the first power is a common error. Example: $\dfrac{5x+4}{(x-1)(x+2)^2}\equiv\dfrac{A}{x-1}+\dfrac{B}{x+2}+\dfrac{C}{(x+2)^2}$, so $5x+4\equiv A(x+2)^2+B(x-1)(x+2)+C(x-1)$. $x=1$: $A=1$. $x=-2$: $-6=-3C$, so $C=2$. Substitution cannot reach $B$ directly, so compare the coefficients of $x^2$: $0=A+B$, so $B=-1$.

A factor such as $(2x-1)$ is handled the same way, with a term $\dfrac{A}{2x-1}$. Substitute $x=\frac12$ to make it zero. Check every answer by substituting a convenient value of $x$ (such as $x=0$) into both sides. At $x=0$, $\dfrac{7x+1}{(x-1)(x+2)}=-\dfrac12$ and $\dfrac{8/3}{-1}+\dfrac{13/3}{2}=-\dfrac12$, which agrees.