The given rational function has a linear factor (x - 1) and an irreducible quadratic factor (x² + 2) in the denominator. Therefore, the decomposition takes the form (A)/(x - 1) + (Bx + C)/(x² + 2), where A, B, and C are real constants to be determined.

The given rational function has a linear factor (x - 1) and an irreducible quadratic factor (x² + 2) in the denominator. Therefore, the decomposition takes the form (A)/(x - 1) + (Bx + C)/(x² + 2), where A, B, and C are real constants to be determined.

Multiplying both sides of the equation by the common denominator (x - 1)(x² + 2) clears the fractions. This gives an algebraic identity true for all values of x.

Substituting x = 1 gives 4(1) + 3 = A(1 + 2) + 0, so 7 = 3A and A = (7)/(3). Equating coefficients of x² gives 0 = A + B, which implies B = -A = -(7)/(3). Setting x = 0 gives 3 = 2A - C, so 3 = 2((7)/(3)) - C = (14)/(3) - C, yielding C = (14)/(3) - 3 = (5)/(3).

Substituting x = 1 gives 4(1) + 3 = A(1 + 2) + 0, so 7 = 3A and A = (7)/(3). Equating coefficients of x² gives 0 = A + B, which implies B = -A = -(7)/(3). Setting x = 0 gives 3 = 2A - C, so 3 = 2((7)/(3)) - C = (14)/(3) - C, yielding C = (14)/(3) - 3 = (5)/(3).

Question 4. Express $\frac{4x + 3}{(x - 1)(x^2 + 2)}$ in partial fractions. Step 1: Set up the decomposition: $\frac{4x + 3}{(x - 1)(x^2 + 2)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 2}$ Step 2: Clear denominators: $4x + 3 = A(x^2 + 2) + (Bx + C)(x - 1)$ Step 3: Solve for constants. Let $x = 1$ : $4(1) + 3 = A((1)^2 + 2) \rightarrow 7 = 3A \rightarrow A = \frac{7}{3}$ Let $x = 0$ : $3 = A(2) + C(-1) \rightarrow 3 = 2\left(\frac{7}{3}\right) - C$ $3 = \frac{14}{3} - C \rightarrow C = \frac{14}{3} - 3 \rightarrow C = \frac{5}{3}$ Compare coefficients of $x^2$ : $0 = A + B \rightarrow 0 = \frac{7}{3} + B \rightarrow B = -\frac{7}{3}$ Step 4: Write the final answer. $\frac{7/3}{x - 1} + \frac{-7/3x + 5/3}{x^2 + 2} \rightarrow \frac{7}{3(x - 1)} + \frac{-7x + 5}{3(x^2 + 2)}$ $\text{Final Answer: } \left[\frac{7}{3(x - 1)} + \frac{5 - 7x}{3(x^2 + 2)}\right]$

Answer: $\frac{7}{3(x - 1)} + \frac{5 - 7x}{3(x^2 + 2)}$

Step-by-step solution

Step 1: Set up the partial fraction decomposition

The given rational function has a linear factor $(x - 1)$ and an irreducible quadratic factor $(x^2 + 2)$ in the denominator. Therefore, the decomposition takes the form $\frac{A}{x - 1} + \frac{Bx + C}{x^2 + 2}$ , where $A$ , $B$ , and $C$ are real constants to be determined.

Step 2: Clear denominators to form an identity

Multiplying both sides of the equation by the common denominator $(x - 1)(x^2 + 2)$ clears the fractions. This gives an algebraic identity true for all values of $x$ .

Step 3: Determine the constants A, B, and C

Substituting $x = 1$ gives $4(1) + 3 = A(1 + 2) + 0$ , so $7 = 3A$ and $A = \frac{7}{3}$ . Equating coefficients of $x^2$ gives $0 = A + B$ , which implies $B = -A = -\frac{7}{3}$ . Setting $x = 0$ gives $3 = 2A - C$ , so $3 = 2\left(\frac{7}{3}\right) - C = \frac{14}{3} - C$ , yielding $C = \frac{14}{3} - 3 = \frac{5}{3}$ .

Step 4: Substitute the constants to obtain the final partial fractions

Substitute the values of $A = \frac{7}{3}$ , $B = -\frac{7}{3}$ , and $C = \frac{5}{3}$ back into the original partial fraction setup. Factoring out $\frac{1}{3}$ simplifies the terms into $\frac{7}{3(x - 1)} + \frac{5 - 7x}{3(x^2 + 2)}$ .

Solve your own maths question free →