Problem 46 - Contemporary Precalculus

Question

In Exercises \(41-46,\) find the constants \(A, B,\) and \(C\). $$\frac{x-2}{(x+4)\left(x^{2}+2 x+2\right)}=\frac{A}{x+4}+\frac{B x+C}{x^{2}+2 x+2}$$

Step-by-Step Solution

Answer

Question: Find the constants A, B, and C such that the following is true: $$\frac{x-2}{(x+4)\left(x^{2}+2 x+2\right)}=\frac{-\frac{3}{5}}{x+4}+\frac{\frac{3}{5} x-\frac{1}{5}}{x^{2}+2 x+2}$$ Answer: A = -3/5, B = 3/5, C = -1/5

Step 1: Multiply both sides by the denominator of the left side

To clear the denominators, we will multiply both sides by \((x+4)(x^2+2x+2)\): \((x-2)=A(x^2+2x+2)+(Bx+C)(x+4)\) Now, we need to find \(A, B,\) and \(C\) such that this equality holds true for all values of \(x\).

Step 2: Expand the right side

Expand the expression on the right side: \(x - 2 = Ax^2 + 2Ax + 2A + Bx^2 + 4Bx + Cx + 4C\) Now, we can group the terms with the same degree of \(x\): \(x - 2 = (A + B)x^2 + (2A + 4B + C)x + (2A + 4C)\)

Step 3: Match coefficients

We will equate coefficients of the left side with the right side for each degree of \(x\): Degree 2: \(0 = A + B\) Degree 1: \(1 = 2A + 4B + C\) Degree 0: \(-2 = 2A + 4C\)

Step 4: Solve the system of equations

We will solve the system of equations using substitution or elimination method: From Degree 2 equation, we get: \(B = -A\) From Degree 0 equation, we can find \(A\): \(-2 = 2A + 4C \implies A = -1 - 2C\) Substitute \(A\) in the equation obtained from Degree 2: \(B = -(-1 - 2C) \implies B = 1 + 2C\) Now, substitute \(A\) and \(B\) in the equation obtained from Degree 1: \(1 = 2(-1 - 2C) + 4(1 + 2C) + C\) Solve for \(C\): \(1 = -2 - 4C + 4 + 8C + C\) \(1 = 2 + 5C\) \(C = -\frac{1}{5}\) Now, substitute \(C\) in the equations for \(A\) and \(B\): \(A = -1 - 2(-\frac{1}{5}) \implies A = -\frac{3}{5}\) \(B = 1 + 2(-\frac{1}{5}) \implies B = \frac{3}{5}\)

Step 5: Write the final decomposition

Now that we have found the values for \(A, B,\) and \(C\), we can write the partial fraction decomposition as: $$\frac{x-2}{(x+4)\left(x^{2}+2 x+2\right)}=\frac{-\frac{3}{5}}{x+4}+\frac{\frac{3}{5} x-\frac{1}{5}}{x^{2}+2 x+2}$$

Key Concepts

Polynomial Long Division

Polynomial long division is a technique used to divide polynomials, similar to how you would perform division with numbers. This method is particularly useful when you have a higher-degree polynomial divided by a lower-degree polynomial, allowing you to separate it into more manageable parts.
To perform polynomial long division, follow these steps:

• Write down the dividend (the polynomial you're dividing) and the divisor (the polynomial you're dividing by).
• Determine how many times the leading term of the divisor can fit into the leading term of the dividend, and this forms the first term of the quotient.
• Multiply this term by the entire divisor, and subtract the result from the original dividend.
• Bring down the next terms of the dividend if necessary, and repeat these steps until you cannot divide any further.

This method allows you to simplify complex polynomial expressions, especially when handling rational functions, ensuring that you can analyze them effectively by breaking them into simpler linear or quadratic factors.

Linear Equations

Linear equations are algebraic expressions where the highest power of the variable is 1. These equations form straight-line graphs when plotted on a coordinate plane. They are fundamental because they are straightforward to solve and form the basis of more complex algebraic operations used in solving systems of equations.
In the context of the problem, linear equations arise when matching coefficients during the partial fraction decomposition. By equating coefficients of corresponding powers of variables, we derive a system of linear equations. These are simple equations like:

• Degree 2: \(0 = A + B\)
• Degree 1: \(1 = 2A + 4B + C\)
• Degree 0: \(-2 = 2A + 4C\)

Solving these equations allows for determining the constants, \(A, B\), and \(C\), needed in partial fractions decomposition. Once found, these constants can be substituted back to provide a simplified expression.

Rational Functions

Rational functions are expressions formed by the ratio of two polynomials, represented generally as \(\frac{P(x)}{Q(x)}\), where both \(P\) and \(Q\) are polynomials. Understanding rational functions is crucial as they appear frequently in various fields of mathematics and applied sciences.
Partial fractions decomposition is an essential concept linked with rational functions. It allows the breaking down of complex rational expressions into simpler fractions that are easier to integrate or analyze. To begin the process:

• Ensure the degree of the numerator is less than the degree of the denominator. If not, perform polynomial long division first.
• Express the rational function as a sum of simpler fractions, aligning each fraction with a factor of the denominator.
• Find constants like \(A, B,\) and \(C\) through methods such as comparing coefficients or substitution.

This breakdown simplifies working with rational functions, particularly in calculus, enabling the tackling of integration and differentiation tasks more effectively by working with simpler components rather than complex fractions.