Use the Factor Theorem: If \( (x + 1) \) is a factor, then \( P(-1) = 0 \). M1

\( P(x) = 3x^3 + 5x^2 + x – 1 \).

Substitute \( x = -1 \):

\( P(-1) = 3(-1)^3 + 5(-1)^2 + (-1) – 1 = 3(-1) + 5(1) – 1 – 1 = -3 + 5 – 1 – 1 = 0 \). A1

Since \( P(-1) = 0 \), \( (x + 1) \) is a factor of \( P(x) \).

[2 marks]

From part (a), \( (x + 1) \) is a factor. Divide \( P(x) = 3x^3 + 5x^2 + x – 1 \) by \( x + 1 \). M1

Using synthetic division with root \( x = -1 \):

Coefficients: \( 3, 5, 1, -1 \).

\( -1 | 3 \ 5 \ 1 \ -1 \)

\( \ \ \ \ \ \ -3 \ -2 \ 1 \)

\( \ \ \ \ 3 \ 2 \ -1 \ 0 \)

Quotient: \( 3x^2 + 2x – 1 \), remainder: 0.

Thus: \( P(x) = (x + 1)(3x^2 + 2x – 1) \).

Factorize \( 3x^2 + 2x – 1 \):

Discriminant: \( \Delta = 2^2 – 4 \cdot 3 \cdot (-1) = 4 + 12 = 16 \).

Roots: \( x = \frac{-2 \pm \sqrt{16}}{6} = \frac{-2 \pm 4}{6} \), so \( x = \frac{1}{3} \), \( x = -1 \).

Thus: \( 3x^2 + 2x – 1 = (x + 1)(3x – 1) \). A1

So: \( P(x) = (x + 1)(x + 1)(3x – 1) = (x + 1)^2 (3x – 1) \). A1

[2 marks]

\( Q(x) = (x + 1)(2x + 1) \).

Express: \( \frac{1}{(x + 1)(2x + 1)} = \frac{A}{x + 1} + \frac{B}{2x + 1} \). M1

Multiply through: \( 1 = A(2x + 1) + B(x + 1) \).

Substitute values:

At \( x = -1 \): \( 1 = A(2(-1) + 1) + B(-1 + 1) = A(-1) \), so \( A = -1 \).

At \( x = -\frac{1}{2} \): \( 1 = A(2(-\frac{1}{2}) + 1) + B(-\frac{1}{2} + 1) = A(0) + B \cdot \frac{1}{2} \), so \( 1 = \frac{B}{2} \), \( B = 2 \). A1

Thus: \( \frac{1}{Q(x)} = -\frac{1}{x + 1} + \frac{2}{2x + 1} \). A1

[2 marks]

\( (x + 1)Q(x) = (x + 1)^2 (2x + 1) \).

Express: \( \frac{1}{(x + 1)^2 (2x + 1)} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2} + \frac{C}{2x + 1} \). M1

Multiply through: \( 1 = A(x + 1)(2x + 1) + B(2x + 1) + C(x + 1)^2 \).

Substitute values:

At \( x = -\frac{1}{2} \): \( 1 = C \left( -\frac{1}{2} + 1 \right)^2 = C \cdot \frac{1}{4} \), so \( C = 4 \).

At \( x = -1 \): \( 1 = B(2(-1) + 1) = B(-1) \), so \( B = -1 \).

Equate coefficients of \( x^2 \): Expand right-hand side:

\( A(2x^2 + 3x + 1) + B(2x + 1) + C(x^2 + 2x + 1) \).

\( x^2 \): \( 2A + C = 0 \). Since \( C = 4 \), \( 2A + 4 = 0 \), \( A = -2 \). A1

Thus: \( \frac{1}{(x + 1)^2 (2x + 1)} = \frac{4}{2x + 1} – \frac{2}{x + 1} – \frac{1}{(x + 1)^2} \). A1

[2 marks]

From part (d): \( \frac{1}{(x + 1)^2 (2x + 1)} = \frac{4}{2x + 1} – \frac{2}{x + 1} – \frac{1}{(x + 1)^2} \). M1

Integrate term by term:

\( \int \left( \frac{4}{2x + 1} – \frac{2}{x + 1} – \frac{1}{(x + 1)^2} \right) dx \).

First term: \( \int \frac{4}{2x + 1} \, dx = 4 \cdot \frac{1}{2} \ln |2x + 1| = 2 \ln |2x + 1| \).

Second term: \( \int -\frac{2}{x + 1} \, dx = -2 \ln |x + 1| \).

Third term: \( \int -\frac{1}{(x + 1)^2} \, dx = -\int (x + 1)^{-2} \, dx = -\cdot \frac{(x + 1)^{-1}}{-1} = \frac{1}{x + 1} \). A1

Combine: \( 2 \ln |2x + 1| – 2 \ln |x + 1| + \frac{1}{x + 1} + C \). A1

[3 marks]

(i) Find \( \lim_{x \to -1} f(x) \).

\( f(x) = \frac{P(x)}{(x + 1)Q(x)} = \frac{3x^3 + 5x^2 + x – 1}{(x + 1)^2 (2x + 1)} \).

From part (b), \( P(x) = (x + 1)^2 (3x – 1) \).

So: \( f(x) = \frac{(x + 1)^2 (3x – 1)}{(x + 1)^2 (2x + 1)} = \frac{3x – 1}{2x + 1} \), for \( x \neq -1 \). M1

Evaluate: \( \lim_{x \to -1} \frac{3x – 1}{2x + 1} = \frac{3(-1) – 1}{2(-1) + 1} = \frac{-3 – 1}{-2 + 1} = \frac{-4}{-1} = 4 \).

(ii) Find \( \lim_{x \to \infty} f(x) \).

Using factored form: \( f(x) = \frac{3x – 1}{2x + 1} \).

As \( x \to \infty \): \( \frac{3x – 1}{2x + 1} \approx \frac{3x}{2x} = \frac{3}{2} \). M1

Alternatively, divide numerator and denominator by \( x^3 \):

\( f(x) = \frac{3x^3 + 5x^2 + x – 1}{x^3 \cdot \frac{(x + 1)^2 (2x + 1)}{x^3}} = \frac{3 + \frac{5}{x} + \frac{1}{x^2} – \frac{1}{x^3}}{\left( 1 + \frac{1}{x} \right)^2 \cdot \left( 2 + \frac{1}{x} \right)} \).

As \( x \to \infty \): \( \frac{3 + 0 + 0 – 0}{(1 + 0)^2 \cdot (2 + 0)} = \frac{3}{1 \cdot 2} = \frac{3}{2} \). A1

Thus, \( \lim_{x \to \infty} f(x) = \frac{3}{2} \). A1

[3 marks]

Total [12 marks]