Given the rational function \( \frac{4 x^{2}+7 x+3}{(x+1)\left(x^{2}+2 x+2\right)} \), perform the following tasks: a) Analyse the given function and identify the types of partial fractions required for decomposition. b) Decompose the function into partial fractions. c) Verify the correctness of your decomposition by combining the partial fractions back into a single rational function.

To analyze the rational function \( \frac{4 x^{2}+7 x+3}{(x+1)\left(x^{2}+2 x+2\right)} \), we will follow the steps outlined in the tasks. ### Part a: Analyze the Function The denominator of the function is \( (x+1)(x^2 + 2x + 2) \). 1. **Identify the factors**: - \( x + 1 \) is a linear factor. - \( x^2 + 2x + 2 \) is a quadratic factor. To check if it can be factored further, we can calculate its discriminant: \[ D = b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot 2 = 4 - 8 = -4 \] Since the discriminant is negative, \( x^2 + 2x + 2 \) does not have real roots and cannot be factored further over the reals. 2. **Types of partial fractions**: - For the linear factor \( x + 1 \), we will have a term of the form \( \frac{A}{x + 1} \). - For the quadratic factor \( x^2 + 2x + 2 \), we will have a term of the form \( \frac{Bx + C}{x^2 + 2x + 2} \). Thus, the partial fraction decomposition will take the form: \[ \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 2x + 2} \] ### Part b: Decompose the Function We set up the equation: \[ \frac{4x^2 + 7x + 3}{(x + 1)(x^2 + 2x + 2)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 2x + 2} \] Multiplying both sides by the denominator \( (x + 1)(x^2 + 2x + 2) \) gives: \[ 4x^2 + 7x + 3 = A(x^2 + 2x + 2) + (Bx + C)(x + 1) \] Expanding the right-hand side: \[ A(x^2 + 2x + 2) + (Bx + C)(x + 1) = A x^2 + 2A x + 2A + Bx^2 + Bx + Cx + C \] \[ = (A + B)x^2 + (2A + B + C)x + (2A + C) \] Now, we equate coefficients from both sides: 1. For \( x^2 \): \( A + B = 4 \) 2. For \( x \): \( 2A + B + C = 7 \) 3. For the constant term: \( 2A + C = 3 \) Now we have a system of equations: 1. \( A + B = 4 \) (1) 2. \( 2A + B + C = 7 \) (2) 3. \( 2A + C = 3 \) (3) From equation (1), we can express \( B \) in terms of \( A \): \[ B = 4 - A \] Substituting \( B \) into equations (2) and (3): From (3): \[ 2A + C = 3 \implies C = 3 - 2A \] Substituting \( B \) and \( C \) into (2): \[ 2A + (4 - A) + (3 - 2A) = 7 \] \[ 2A + 4 - A + 3 - 2A = 7 \] \[ 4 - A = 7 \implies -A = 3 \implies A = -3 \] Now substituting \( A = -3 \) back to find \( B \) and \( C \): \[ B = 4 - (-3) = 7 \] \[ C = 3 - 2(-3) = 3 + 6 = 9 \] Thus, we have: \[ A = -3, \quad B = 7, \quad C = 9 \] The partial fraction decomposition is: \[ \frac{-3}{x + 1} + \frac{7x + 9}{