Question

If \(\frac{\mathrm{x}^{3}}{(\mathrm{x}-1)(\mathrm{x}-2)}=\mathrm{Ax}+\mathrm{B}+\frac{\mathrm{C}}{\mathrm{x}-1}+\frac{\mathrm{D}}{\mathrm{x}-2}\), then \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) and \(\mathrm{D}\) respectively are (1) \(1, \quad 3,1,8\) (2) \(1,-1,3,8\) (3) \(1, \quad 3,-1, \quad 8\) (4) \(-1,-3,1,8\)

Short answer

Question: Given the equation \(\frac{x^3}{(x-1)(x-2)} = Ax+B+\frac{C}{x-1}+\frac{D}{x-2}\), find the values of A, B, C, and D. Answer: A = 1, B = -1, C = 3, D = 8

Step-by-step solution

Clear denominators by multiplying

To clear denominators, multiply both sides of the equation by the common denominator (\((x-1)(x-2)\)): \((x-1)(x-2)\biggr(\frac{x^3}{(x-1)(x-2)}\biggr)=(x-1)(x-2)(Ax+B+\frac{C}{x-1}+\frac{D}{x-2})\) This should give us a polynomial equation matching the left side with the right side.

Simplify both sides

After multiplying, we simplify both sides: \(x^3 = A(x-1)(x-2) + B(x-1)(x-2) + C(x-2) + D(x-1)\)

Expand the equation

We expand the equation to get: \(x^3 = A(x^2 - 3x + 2) + B(x^2 - 3x + 2) + Cx - C2 + Dx - D\)

Combine like terms

Now, combine the like terms on the right-hand side: \(x^3 = (A+B)x^2 + (-3A-3B+C+D)x + (2A-2B-2C+D)\)

Get conditions from coefficients

Find the conditions by equalizing the coefficients of like terms: \(1. \ Coefficient\ of\ x^2: 0 = A+B\) \(2. \ Coefficient\ of\ x: 0 = -3A -3B + C + D\) \(3. \ Coefficient\ of\ constant\ term: 1 = 2A - 2B - 2C + D\)

Solve system of equations

Solve the system of equations obtained in step 5: Using the first condition (\(A+B=0\)), we find that \(A=-B\). Substituting this in the second condition, we get: \(0 = -3(-B) -3B + C + D => C + D = 6B\). Substituting A and B values in the third condition, we get: \(1 = 2(-B) - 2B - 2C + D => D - 2C = 4B - 1\). Now, we have two equations in C and D: (i) \(D-2C=4B-1\) (ii) \(D=6B-C\) Solving (i) and (ii), we get \(C=3\) and \(D=8\). Using these values, from the first and second conditions, we find that \(A=1\) and \(B=-1\). Therefore, the correct option is (2): \(A=1, B=-1, C=3,\) and \(D=8\).

Key concepts

Polynomial Equations

Polynomial equations are mathematical expressions consisting of variables and coefficients connected by addition, subtraction, multiplication, and non-negative integer exponentiation of the variables. When solving polynomial equations, the goal is often to find the values of the variables that will make the equation true.

For example, in the given exercise, \(x^3\) is a polynomial equation because it involves a variable (\(x\)) raised to a positive integer power (3) without division by a variable. In the process of partial fraction decomposition, we often encounter a scenario where, after clearing the denominators, we are left with a polynomial equation. The subsequent task is to find coefficients that satisfy the equation for all values of \(x\).

To simplify the process, we equate coefficients of the same degree on both sides of the equation. This creates a system of linear equations that we can solve for the unknown coefficients, which are essential to complete the partial fraction decomposition. Illustrating this concept with clarity is fundamental for students to understand how to manipulate and solve polynomial equations.

Systems of Linear Equations

A system of linear equations consists of two or more equations that share a set of common variables. The solution to such a system is the set of values that satisfy all equations in the system simultaneously. There are various methods to solve these systems, such as graphing, substitution, elimination, and matrix operations.

In our exercise, the partial fraction decomposition leads to a system of linear equations derived from equating the coefficients of like terms in the expanded polynomial expression. Solving the system allows us to determine the unknown coefficients \(A\), \(B\), \(C\), and \(D\).

Understanding how to manipulate and solve systems of linear equations is crucial when approaching problems like the one in the exercise, especially when dealing with polynomial equations during the process of partial fraction decomposition. Breaking down this system into simpler equations, step by step, ensures that students can follow the solution process and grasp the underlying algebraic principles.

Algebraic Fractions

Algebraic fractions are fractions in which the numerator or the denominator (or both) contains variables. Like numerical fractions, algebraic fractions can be simplified, added, subtracted, multiplied, and divided. Decomposing complex algebraic fractions into simpler parts often involves the technique of partial fraction decomposition.

In our exercise, the given algebraic fraction \(\frac{x^3}{(x-1)(x-2)}\) is broken down into a sum of simpler fractions with undetermined coefficients \(A\), \(B\), \(C\), and \(D\). The process of finding these coefficients requires clearing the denominator and equating the resultant polynomial to a target polynomial obtained through the proposed decomposition.

Presenting algebraic fractions in a relatable way and connecting them with the process of partial fraction decomposition can enhance student comprehension. It is important for educational content to guide students through these complex processes in a step-wise and understandable manner, ensuring they gain the competency to work with algebraic fractions confidently.