Partial fractions practice questions.

It is given that (7x + 1)/((x − 1)(x + 3)) ≡ A/(x − 1) + B/(x + 3). Determine the value of B.

• 2
• 5
• −5
• 7

Answer: 5

Multiplying out gives 7x + 1 ≡ A(x + 3) + B(x − 1); substituting x = −3 gives −20 = −4B, so B = 5 (and x = 1 gives A = 2).

Which of the following is the partial fraction decomposition of (x + 7)/(x² − x − 6)?

• 2/(x − 3) − 1/(x + 2)
• 2/(x − 3) + 1/(x + 2)
• 1/(x − 3) − 2/(x + 2)
• 2/(x + 2) − 1/(x − 3)

Answer: 2/(x − 3) − 1/(x + 2)

Factorising gives (x − 3)(x + 2); then x + 7 ≡ A(x + 2) + B(x − 3), so x = 3 gives A = 2 and x = −2 gives B = −1.

Using the decomposition 1/(r(r + 1)) ≡ 1/r − 1/(r + 1), calculate the sum from r = 1 to r = 10 of 1/(r(r + 1)).

• 9/10
• 11/12
• 1
• 10/11

Answer: 10/11

The sum telescopes: all middle terms cancel, leaving 1/1 − 1/11 = 10/11.

Given that (5x − 7)/((x − 1)(x − 2)) ≡ A/(x − 1) + B/(x − 2), calculate the value of A + B.

• −7
• 6
• 5
• 1

Answer: 5

Substituting x = 1 gives −2 = −A so A = 2; substituting x = 2 gives B = 3; hence A + B = 5, which must equal the coefficient of x in the numerator.

To decompose (2x + 1)/((x − 3)(x + 5)) ≡ A/(x − 3) + B/(x + 5), both sides are multiplied by (x − 3)(x + 5). Which value of x should then be substituted to find A directly?

• x = −5
• x = 3
• x = −3
• x = 5

Answer: x = 3

Substituting x = 3 makes the factor (x − 3) zero, eliminating the B term and leaving an equation in A alone.

Before writing (x² + 3x + 5)/((x + 1)(x + 2)) in partial fractions, what must be done first?

• Complete the square in the numerator
• Substitute x = −1 into the numerator
• Divide, since deg(numerator) ≥ deg(denominator)
• Multiply the numerator and denominator by (x + 1)(x + 2)

Answer: Divide, since deg(numerator) ≥ deg(denominator)

Partial fraction decomposition requires a proper fraction; since numerator and denominator both have degree 2, divide first to obtain 1 + (remainder)/((x + 1)(x + 2)).