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Chapter 4: Problem 217

In Problem 216 you see that when we added numerical multiples of the reciprocals of first degree polynomials we got a fraction in which the denominator is a quadratic polynomial. This will always happen unless the two denominators are multiples of each other, because their least common multiple will simply be their product, a quadratic polynomial. This leads us to ask whether a fraction whose denominator is a quadratic polynomial can always be expressed as a sum of fractions whose denominators are first degree polynomials. Find numbers \(c\) and \(d\) so that $$ \frac{5 x+1}{(x-3)(x+5)}=\frac{c}{x-3}+\frac{d}{x+5} $$

Short Answer

Expert verified
The values are \(c = 2\) and \(d = 3\).

Step by step solution

01

Set up the equation

We start with the given equation:\[ \frac{5x + 1}{(x-3)(x+5)} = \frac{c}{x-3} + \frac{d}{x+5} \]The goal is to find values of \(c\) and \(d\) that satisfy this equation.
02

Combine the right-hand side

To combine the fractions on the right-hand side, we need a common denominator:\[ \frac{c}{x-3} + \frac{d}{x+5} = \frac{c(x+5) + d(x-3)}{(x-3)(x+5)} \]
03

Set the numerators equal

Since the denominators are equal, set the numerators equal:\[ 5x + 1 = c(x + 5) + d(x - 3) \]
04

Distribute and simplify

Expand the terms on the right side:\[ 5x + 1 = c x + 5c + d x - 3d \]Combine like terms:\[ 5x + 1 = (c + d)x + (5c - 3d) \]
05

Match coefficients

Match the coefficients of like terms on both sides of the equation. For the coefficients of \(x\):\[ 5 = c + d \]For the constant terms:\[ 1 = 5c - 3d \]
06

Solve the system of equations

Solve the system of equations:\[ \begin{cases} 5 = c + d \ 1 = 5c - 3d \end{cases} \]First, solve for \(d\):\[ d = 5 - c \]Substitute into the second equation:\[ 1 = 5c - 3(5 - c) \]Simplify and solve for \(c\):\[ 1 = 5c - 15 + 3c \]\[ 1 = 8c - 15 \]\[ 16 = 8c \]\[ c = 2 \]Now, find \(d\):\[ d = 5 - c = 5 - 2 = 3 \]
07

Verify the solution

Substitute \(c = 2\) and \(d = 3\) back into the original equation to verify:\[ \frac{5x + 1}{(x-3)(x+5)} = \frac{2}{x-3} + \frac{3}{x+5} \]Combine the right-hand side to check if it matches:\[ \frac{2(x+5) + 3(x-3)}{(x-3)(x+5)} = \frac{2x + 10 + 3x - 9}{(x-3)(x+5)} = \frac{5x + 1}{(x-3)(x+5)} \]The equation holds true, so the values found are correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rational Expressions
Rational expressions are fractions where the numerator and the denominator are polynomials. These expressions often come up in algebra and calculus. Understanding how to manipulate and simplify rational expressions is vital.

For example, in the given problem, we dealt with the expression \(\frac{5x + 1}{(x-3)(x+5)}\). Here, both the numerator \(5x + 1\) and the denominator \((x-3)(x+5)\) are polynomials. Simplifying this expression or breaking it down into simpler parts makes it easier to work with.
Algebraic Fractions
Algebraic fractions are similar to rational expressions but specifically emphasize fractions that contain algebraic expressions in either the numerator or the denominator.

A key technique in our problem was partial fraction decomposition, which writes a complex algebraic fraction as a sum of simpler fractions. For instance, from \(\frac{5x + 1}{(x-3)(x+5)}\), we aim to express it as \(\frac{c}{x-3} + \frac{d}{x+5}\). By matching coefficients, we simplified the complex fraction into more manageable parts.
Quadratic Polynomials
A quadratic polynomial is a second-degree polynomial, typically in the form \(ax^2 + bx + c\). In our exercise, \( (x-3)(x+5) \) serves as the quadratic polynomial in the denominator.

We decomposed the fraction by recognizing that the denominator was a product of two first-degree polynomials. This allowed us to express the complex fraction as a sum of simpler fractions, each with a first-degree polynomial in the denominator.
System of Equations
A system of equations is a set of two or more equations with the same variables. To find values for these variables, you need to solve the system. In our problem, after combining the right-hand side into a single fraction, we set the numerators equal creating a system of equations:
\[ 5 = c + d \] and \[ 1 = 5c - 3d \].
By solving this system, we found the values of \c\ and \d\. This approach allowed us to break down the more complex expression into simpler components.

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