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P4.2 - Algebra and functions

Syllabus
2019
Topic
P4.2
Level
A2

Learning objectives

P4.2.1Decompose rational functions into Partial fractionsDecompose rational functions into Partial fractions to include denominators such as partial fractions (denominators not (ax + b)(cx + d)(ex + f) and (ax + b)(cx + d)2. more complicated than repeated linear terms).; The degree of the numerator may equal or exceed the degree of the denominator.; Applications to integration, differentiation and series expansions.; Quadratic factors in the denominator such as (x2 + a), a > 0, are not required.

Choose and find a partial-fraction form

Partial fractions rewrite one rational function as a sum of simpler rational terms. The denominator determines the complete template; once the coefficients are found, the equivalent form is easier to integrate, differentiate or expand as a series.

Denominator or degree Required starting form
(ax+b)(cx+d)(ex+f)(ax+b)(cx+d)(ex+f) Aax+b+Bcx+d+Cex+f\frac{A}{ax+b}+\frac{B}{cx+d}+\frac{C}{ex+f}
(ax+b)(cx+d)2(ax+b)(cx+d)^2 Aax+b+Bcx+d+C(cx+d)2\frac{A}{ax+b}+\frac{B}{cx+d}+\frac{C}{(cx+d)^2}
numerator degree ≥\ge denominator degree divide first: polynomial quotient + proper remainder

Factor the denominator, divide first if the fraction is improper, and write every required term. Multiply through by the full denominator to form a polynomial identity. Substitute roots of the linear factors to isolate coefficients, then compare coefficients or use another convenient value for any coefficient left. Recombine the result over the original denominator to check it.

For example, 5x+1(x−1)(x+2)2=Ax−1+Bx+2+C(x+2)2.\frac{5x+1}{(x-1)(x+2)^2}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}. Multiplying through gives 5x+1=A(x+2)2+B(x−1)(x+2)+C(x−1).5x+1=A(x+2)^2+B(x-1)(x+2)+C(x-1). Setting x=1x=1 gives A=2/3A=2/3; setting x=−2x=-2 gives C=3C=3; comparing x2x^2 coefficients gives B=−2/3B=-2/3.

Next operation Why the split helps
Integration linear-denominator terms give logarithms; repeated terms use a power rule
Differentiation simple reciprocal powers differentiate term by term
Series expansion rewrite terms into (1+ux)n(1+ux)^n form, retaining each convergence condition

A repeated factor needs a term for every power up to its multiplicity; omitting one makes the identity incomplete. Preserve values excluded by the original denominator. Denominators with irreducible quadratic factors such as x2+ax^2+a, a>0a>0, are outside this specification.