Questions on partial fractions

Question 1

Express (3x + 5)/((x + 1)(x + 2)) in partial fractions.

Accepted Answer

Set the fraction equal to A/(x + 1) + B/(x + 2) and multiply up, giving 3x + 5 = A(x + 2) + B(x + 1). Substituting x = −1 kills the B bracket and leaves 2 = A. Substituting x = −2 leaves −1 = −B, so B = 1. The answer is 2/(x + 1) + 1/(x + 2). B1 for the correct form, M1 for the substitutions and A1 for the two constants, so the correct form is credited even when the constants are not found.

Question 2

Writing (3x + 5)/((x + 1)(x + 2)) as A/(x + 1) + B/(x + 2) gives 3x + 5 = A(x + 2) + B(x + 1). Explain why substituting x = −1 finds A immediately.

Accepted Answer

At x = −1 the factor (x + 1) is zero, so the whole B term vanishes and the equation collapses to a statement about A alone. Each root of the denominator switches off every constant except its own. B1 B1 for the vanishing factor and for the equation in A alone. Comparing coefficients reaches the same values, but it needs simultaneous equations where substitution needs none.

Question 3

Write down the form of the partial fractions for (5x − 1)/((x − 2)(x + 3)²), using constants A, B and C, without finding the constants.

Accepted Answer

The distinct bracket takes one term and the squared bracket takes two, one over the bracket and one over its square: A/(x − 2) + B/(x + 3) + C/(x + 3)². B1 for the term over (x − 2), B1 for the pair over (x + 3) and (x + 3)². Three constants for a cubic denominator. A single C/(x + 3)² cannot reproduce a general linear numerator, and (Bx + C)/(x + 3)² is a correct but non-standard alternative that hides the B/(x + 3) term inside it.

Question 4

The cover-up rule finds the constant over a linear bracket by covering that bracket in the denominator and substituting the bracket's root into what remains. Use it to find the constants A and B for which 4/((x − 3)(x + 1)) = A/(x − 3) + B/(x + 1).

Accepted Answer

For A, cover (x − 3) and put x = 3 into what is left: A = 4/(3 + 1) = 1. For B, cover (x + 1) and put x = −1 into what is left: B = 4/(−1 − 3) = −1. M1 for covering a bracket and substituting its root into the rest, A1 for A = 1, A1 for B = −1. So 4/((x − 3)(x + 1)) = 1/(x − 3) − 1/(x + 1). The rule is substitution with the multiplying-up step skipped: 4 = A(x + 1) + B(x − 3) at x = 3 gives 4 = 4A, which is exactly what covering (x − 3) reads off. It works for the constant over a distinct linear bracket; with a repeated bracket it finds the constant over the square but not the one over its first power, and it says nothing about an improper fraction. At x = 0 both sides give −4/3, which checks the split.

Question 5

Express (6x² + 5x − 2)/(x(x − 1)(x + 2)) in partial fractions.

Accepted Answer

Set A/x + B/(x − 1) + C/(x + 2) and multiply up: 6x² + 5x − 2 = A(x − 1)(x + 2) + Bx(x + 2) + Cx(x − 1). Then x = 0 gives −2 = −2A, so A = 1; x = 1 gives 9 = 3B, so B = 3; x = −2 gives 12 = 6C, so C = 2. The answer is 1/x + 3/(x − 1) + 2/(x + 2), and any spare x value, say x = 2, checks it: 32 = 4 + 24 + 4. B1 for the form, M1 for the substitutions, A1 for A = 1 and B = 3, A1 for C = 2.

Question 6

Express (2x + 7)/(x + 3)² in partial fractions.

Accepted Answer

A squared bracket needs both powers, so try A/(x + 3) + B/(x + 3)². Multiplying up gives 2x + 7 = A(x + 3) + B. Substituting x = −3 leaves 1 = B, and comparing the x coefficients gives A = 2. The answer is 2/(x + 3) + 1/(x + 3)². B1 for including both powers, M1 for multiplying up, A1 for B = 1, A1 for A = 2. One fraction per bracket fails here, because a single A/(x + 3)² cannot produce the 2x on the left.

Question 7

Express (x² + 4x + 5)/((x + 1)(x + 2)) in partial fractions.

Accepted Answer

The fraction is improper: the top's degree matches the bottom's. Divide first: x² + 4x + 5 = (x² + 3x + 2) + (x + 3), so the fraction is 1 + (x + 3)/((x + 1)(x + 2)). Splitting the proper part: x = −1 gives A = 2, x = −2 gives B = −1. The answer is 1 + 2/(x + 1) − 1/(x + 2). M1 for spotting the improper fraction and dividing, A1 for the quotient and proper remainder, B1 for the partial-fraction form, A1 for A = 2, A1 for B = −1.

Question 8

Express (7x + 7)/(2x² + 5x − 3) in partial fractions.

Accepted Answer

The denominator arrives unfactorised, so no template can be written until it is in brackets. Factorising, 2x² + 5x − 3 = (2x − 1)(x + 3), and the form is A/(2x − 1) + B/(x + 3). Multiplying up gives 7x + 7 = A(x + 3) + B(2x − 1). Substituting x = −3 gives −14 = −7B, so B = 2. Substituting x = ½ gives 21/2 = (7/2)A, so A = 3. The answer is 3/(2x − 1) + 2/(x + 3). M1 for factorising the denominator, B1 for the correct form, M1 for substituting each bracket's root, A1 for A = 3 and B = 2. The root of a non-monic bracket is a fraction rather than a whole number: 2x − 1 vanishes at x = ½, and substituting x = 1 there silences nothing and leaves two unknowns in one equation. Note also that the numerator over a linear bracket is a constant even when that bracket is non-monic, so A/(2x − 1) is right and (Ax + B)/(2x − 1) is not. At x = 0 both sides give −7/3, which checks the split.

Question 9

Express (5x + 4)/((x − 1)(x + 2)²) in partial fractions.

Accepted Answer

Three terms: A/(x − 1) + B/(x + 2) + C/(x + 2)². B1 for the template with both powers of the repeated bracket. Multiplying up: 5x + 4 = A(x + 2)² + B(x − 1)(x + 2) + C(x − 1). Substituting x = 1 gives 9 = 9A, so A = 1; substituting x = −2 gives −6 = −3C, so C = 2. M1 for the two substitutions, A1 for A = 1 and C = 2. No value of x isolates B, so compare the x² coefficients: the left side has none, and the right has A + B, so B = −1. A1 for B = −1. The answer is 1/(x − 1) − 1/(x + 2) + 2/(x + 2)². Checking the constant terms, 4A − 2B − C = 4 + 2 − 2 = 4, agrees with the left side. The B term is the one that is omitted when the squared bracket is given a single fraction, and it comes out negative here, which is not a sign of error.

Question 10

Given that (px + 7)/((x + 1)(x + 4)) = 2/(x + 1) + q/(x + 4) for all x other than −1 and −4, find the values of the constants p and q.

Accepted Answer

Multiply up: px + 7 = 2(x + 4) + q(x + 1). M1 for the identity. Substituting x = −1 removes the q term: −p + 7 = 2 × 3 = 6, so p = 1. M1 for a substitution that isolates one unknown, A1 for p = 1. Substituting x = −4 removes the 2 term: −4p + 7 = −3q, and with p = 1 that is 3 = −3q, so q = −1. A1. Comparing x coefficients confirms it: p = 2 + q, and 1 = 2 − 1. The decomposition is (x + 7)/((x + 1)(x + 4)) = 2/(x + 1) − 1/(x + 4), and x = 0 gives 7/4 on the left and 2 − 1/4 on the right.

Question 11

By comparing coefficients, find the constants A and B such that (7x − 2)/((x − 2)(x + 1)) = A/(x − 2) + B/(x + 1).

Accepted Answer

Multiply up: 7x − 2 = A(x + 1) + B(x − 2) = (A + B)x + (A − 2B). M1 for expanding and collecting the right side. Two identical polynomials have identical coefficients, so A + B = 7 and A − 2B = −2. A1 for the pair of equations. Subtracting, 3B = 9, so B = 3 and then A = 4. A1 for A = 4 and B = 3. The decomposition is 4/(x − 2) + 3/(x + 1). Substituting x = 2 and x = −1 reaches the same two constants in one line each, and the two methods agree because the identity holds for every x, so any two values of x, or any two coefficients, are enough to fix two unknowns. Comparing coefficients is the method to keep for the constant that no substitution isolates, as happens with a repeated bracket.

Question 12

Express (3x² + 7x + 8)/((x + 1)²(x + 2)) in partial fractions.

Accepted Answer

Three constants: A/(x + 1) + B/(x + 1)² + C/(x + 2). Multiplying up: 3x² + 7x + 8 = A(x + 1)(x + 2) + B(x + 2) + C(x + 1)². Then x = −1 gives 4 = B; x = −2 gives 6 = C; comparing x² coefficients, A + C = 3, so A = −3. The answer is −3/(x + 1) + 4/(x + 1)² + 6/(x + 2). B1 for the three-term form, M1 for multiplying up, A1 for B = 4, A1 for C = 6, M1 for comparing x² coefficients, A1 for A = −3. The check at x = 0 confirms it: 8 = −6 + 8 + 6.

Question 13

Express (2x³ + 7x² + 5x + 4)/((x + 1)(x + 3)) in partial fractions.

Accepted Answer

Compare degrees first. The numerator has degree 3 and the denominator degree 2, so the fraction is improper and the quotient of the division will be linear, not a constant. Dividing by x² + 4x + 3 gives 2x³ + 7x² + 5x + 4 = (x² + 4x + 3)(2x − 1) + 3x + 7, so the fraction is 2x − 1 + (3x + 7)/((x + 1)(x + 3)). Now split the proper remainder: 3x + 7 = A(x + 3) + B(x + 1), and x = −1 gives 4 = 2A, so A = 2, while x = −3 gives −2 = −2B, so B = 1. The answer is 2x − 1 + 2/(x + 1) + 1/(x + 3). M1 for comparing degrees and dividing, A1 for the quotient 2x − 1, A1 for the remainder 3x + 7, B1 for the partial-fraction form of the remainder, A1 for A = 2, A1 for B = 1. At x = 0 the left side is 4/3 and the right side is −1 + 2 + 1/3, which agree.

Question 14

Express f(x) = (5x + 5)/((x − 2)(x + 3)) in partial fractions, and hence show that f has no stationary points.

Accepted Answer

Set A/(x − 2) + B/(x + 3) and multiply up: 5x + 5 = A(x + 3) + B(x − 2). Substituting x = 2 gives 15 = 5A, so A = 3; substituting x = −3 gives −10 = −5B, so B = 2. So f(x) = 3/(x − 2) + 2/(x + 3), which is 3(x − 2)⁻¹ + 2(x + 3)⁻¹. Each piece now differentiates by the chain rule on a negative power, giving f′(x) = −3/(x − 2)² − 2/(x + 3)². Both denominators are squares, so both terms are negative wherever f is defined, their sum is never zero, and f has no stationary points. B1 for the correct form, M1 for multiplying up and substituting the roots, A1 for A = 3 and B = 2, M1 for differentiating the split as two negative powers, A1 for f′(x) with the sign argument. Differentiating the original fraction needs the quotient rule and a numerator that has to be expanded before it can be examined; the split replaces that with two standard derivatives, and the sign of each is readable at sight. Note that f is decreasing on each of the three intervals of its domain separately, since x = 2 and x = −3 are excluded, and a curve in three pieces is not decreasing across the gaps.

Question 15

Explain why (x² + 5x + 7)/(x + 2)² cannot be written in the form A/(x + 2) + B/(x + 2)² for constants A and B, and express it in partial fractions with a suitable form.

Accepted Answer

The fraction is improper: the numerator has degree 2, the same as the denominator. As |x| grows, A/(x + 2) + B/(x + 2)² tends to 0, while the given fraction tends to 1, the ratio of the leading coefficients, so no choice of A and B can make them equal. B1 for the reason. Divide first. Since (x + 2)² = x² + 4x + 4, the numerator is (x² + 4x + 4) + (x + 3), so the fraction is 1 + (x + 3)/(x + 2)². M1 for the division, A1 for the quotient 1 and remainder x + 3. The proper remainder takes the two-term template: (x + 3)/(x + 2)² = A/(x + 2) + B/(x + 2)², so x + 3 = A(x + 2) + B; x = −2 gives B = 1, and comparing x coefficients gives A = 1. M1 for splitting the remainder. The answer is 1 + 1/(x + 2) + 1/(x + 2)². A1. Writing x + 3 as (x + 2) + 1 reaches the same split by inspection. At x = 0 the original is 7/4 and the split gives 1 + 1/2 + 1/4, which agree.