How do I resolve into partial fractions when the denominator has a repeated linear factor in Inter 2nd year Maths IIA?

Question

TS Inter second year, Maths IIA, chapter Partial Fractions. Simple distinct linear factors I can do. But when the denominator has something like (x + 1) squared, I don't know how many fractions to write or how to find the constants quickly.

GradeHunt_Team · Answer

Rule: a repeated linear factor (ax + b) raised to the power k needs k separate fractions, one for each power:
A1/(ax + b) + A2/(ax + b)^2 + ... + Ak/(ax + b)^k

Before you start, check that the degree of the numerator is less than the degree of the denominator. If not, divide first and resolve only the remainder part.

Worked example: resolve (x + 3) / [(x - 1)(x + 1)^2] into partial fractions.

Step 1: write the form. (x + 1) is repeated twice, so it needs two fractions:
(x + 3) / [(x - 1)(x + 1)^2] = A/(x - 1) + B/(x + 1) + C/(x + 1)^2

Step 2: multiply both sides by (x - 1)(x + 1)^2:
x + 3 = A(x + 1)^2 + B(x - 1)(x + 1) + C(x - 1)

Step 3: put convenient values of x.
x = 1: 4 = A(2)^2, so A = 1
x = -1: 2 = C(-2), so C = -1

Step 4: the roots x = 1 and x = -1 do not give B, so compare coefficients of x^2 (putting any other value, such as x = 2, also works).
Left side has 0x^2. Right side has A + B.
0 = A + B, so B = -1

Answer:
(x + 3) / [(x - 1)(x + 1)^2] = 1/(x - 1) - 1/(x + 1) - 1/(x + 1)^2

Step 5: check with x = 0.
Left side: 3 / [(-1)(1)] = -3
Right side: 1/(-1) - 1/1 - 1/1 = -1 - 1 - 1 = -3. It matches.

Common slips:
1. Writing only C/(x + 1)^2 and leaving out B/(x + 1). Every power from 1 to k needs its own fraction.
2. Getting stuck once the root values are used up. Then compare the coefficient of the highest power (or the constant term), or put any other simple value such as x = 2.
3. Skipping the check. Substituting x = 0 or x = 2 takes ten seconds and catches sign errors.

How do I resolve into partial fractions when the denominator has a repeated linear factor in Inter 2nd year Maths IIA?

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