Problem 17 - Precalculus

Question

Determine the constants (denoted by capital letters) so that each equation is an identity. For Exercises 1– 6, do each problem in two ways: (a) use the equating-the-coefficients theorem, as in Example 1; and (b) use the convenient-values method that was explained after Example 1. For the remainder of the exercises, use either method (or a combination). $$\frac{3 x^{2}-2}{(x-2)(x+1)(x-1)}=\frac{A}{x-2}+\frac{B}{x+1}+\frac{C}{x-1}$$

Step-by-Step Solution

Answer

A = \(\frac{4}{3}\), B = \(\frac{1}{6}\), C = -\(\frac{1}{2}\).

Step 1: Clear Denominator

Multiply both sides of the equation by \((x-2)(x+1)(x-1)\) to eliminate the fractions. This gives you: \[ 3x^2 - 2 = A(x+1)(x-1) + B(x-2)(x-1) + C(x-2)(x+1) \]

Step 2: Expand the Right Side

Expand the right side of the equation: - For the term \(A(x+1)(x-1)\), use the distributive property: \[ A(x^2 - 1) = Ax^2 - A \]- For the term \(B(x-2)(x-1)\), use the distributive property: \[ B(x^2 - 3x + 2) = Bx^2 - 3Bx + 2B \]- For the term \(C(x-2)(x+1)\), use the distributive property: \[ C(x^2 - x - 2) = Cx^2 - Cx - 2C \]Combine these to get: \[ (A + B + C)x^2 + (-3B - C)x + (-A + 2B - 2C) \]

Step 3: Equate Coefficients

Since \(3x^2 - 2\) must equal the expanded form of the right side for all \(x\), equate the coefficients of corresponding powers of \(x\):- For \(x^2\) terms, equate \(3\) to \(A + B + C\).- For \(x\) terms, equate \(0\) to \(-3B - C\).- For constant terms, equate \(-2\) to \(-A + 2B - 2C\). This gives the system of equations: 1. \( A + B + C = 3 \)2. \( -3B - C = 0 \)3. \( -A + 2B - 2C = -2 \)

Step 4: Solve the System

Solve the system of equations step by step:- From equation 2: \(-3B = C\) or \(C = -3B\).- Substitute \(C = -3B\) into equation 1: \[ A + B - 3B = 3 \Rightarrow A - 2B = 3 \]- Substitute \(C = -3B\) into equation 3: \[ -A + 2B + 6B = -2 \Rightarrow -A + 8B = -2 \] - Solve for \(A\) and \(B\) using these reduced equations: - From \(-A + 8B = -2\), rearrange to get \(A = 8B + 2\). - Substitute \(A = 8B + 2\) into \(A - 2B = 3\): \[ 8B + 2 - 2B = 3 \Rightarrow 6B = 1 \Rightarrow B = \frac{1}{6} \] - Substitute \(B = \frac{1}{6}\) back into \(A = 8B + 2\): \[ A = 8(\frac{1}{6}) + 2 \Rightarrow A = \frac{4}{3} \] - Now substitute \(B = \frac{1}{6}\) into \(C = -3B\): \[ C = -3(\frac{1}{6}) = -\frac{1}{2} \]

Step 5: Use Convenient Values Method

Use the convenient-values method by setting \(x\) to specific values to find \(A\), \(B\), and \(C\):- Set \(x = 2\), then the left-hand side becomes \(3(2)^2 - 2 = 10\) and the right-hand side becomes \(A(0) = 0\): - Resulting in: \[ 10 = \frac{B}{3} \Rightarrow B = \frac{1}{6} \]- Set \(x = -1\), the left-hand side equals \(0 - 2 = -2\), the right-hand side becomes \(B(0) = 0\): - Resulting in: \[ -2 = \frac{C}{-1} \Rightarrow C = -\frac{1}{2} \]- Set \(x = 1\), the left-hand side becomes \( 3(1)^2 - 2 = 1\), the right-hand side becomes \(C(0) = 0\): - Resulting in: \[ 1 = A / -1 \Rightarrow A = \frac{4}{3} \]

Step 6: Conclusion: Values of A, B, C

By both methods, the values agree: - \(A = \frac{4}{3}\)- \(B = \frac{1}{6}\)- \(C = -\frac{1}{2}\)

Key Concepts

Equating the Coefficients Theorem

The Equating the Coefficients Theorem is a powerful tool for solving polynomial equations when they are written as identities. When you equate two polynomials, ensuring that they are equal for every value of the variable, you rely on the equality of each term's coefficients. This means that the coefficients in front of corresponding powers of the variable on both sides of the equation must be equal.

In our exercise, after eliminating the denominators and expanding the expression on the right side, we set coefficients of similar powers of \(x\) on both sides to be equal.

• Identify corresponding terms: Match the terms of the same power from each side. In our exercise, these were \(x^2\), \(x\), and the constant terms.
• Write equations for each: Form a system of equations where the coefficients of \(x^2\), \(x\), and constant terms are set equal to each other.
• Solve the system: Use these equations to solve for the unknowns \(A\), \(B\), and \(C\).

By applying the equating coefficients theorem accurately, you can methodically determine unknown constants in partial fraction decompositions and other polynomial equations.

Convenient Values Method

The Convenient Values Method simplifies finding unknown constants by strategically choosing values for \(x\) that make parts of the equation zero. This often results in straightforward equations that directly solve for one of the constants, which is easier and faster in many cases than solving simultaneous equations.

Here's how it helped in our example:

• Choose smart values for \(x\): Set \(x\) to critical points (like the roots of the factors in the denominator, which in this case were 2, -1, and 1). These choices simplify one or more terms to zero, isolating a single variable.
• Solve each resulting equation: By using values where one expression becomes zero, you're left with a single expression in terms of one unknown.

This method is especially useful when dealing with partial fraction decompositions where choosing the roots of the denominator maximizes simplification. It's a fast way to validate or derive the values found using other methods like equating coefficients.

Polynomial Expansion

Polynomial Expansion involves rewriting expressions in their expanded form by applying distributive properties and simplifying. This step allows us to see the polynomial in terms of standard powers of \(x\), making it easier to compare and analyze.

In the exercise, we expanded each term of the right-hand side equation after clearing the fractions:

• Distribute through terms: Each binomial in the denominator was multiplied through by its counterparts, simplifying the product to a standard polynomial form.
• Combine like terms: After distributing, gather terms with similar powers of \(x\) together so you could directly equate corresponding coefficients in later steps.

Mastering polynomial expansion is crucial when solving problems involving fractions, partial fractions or complex polynomials, ensuring the expressions are fully interpreted and correctly setup for further steps such as equating coefficients.