Problem 12 - Precalculus: Functions and Graphs

Question

Find the partial fraction decomposition. \(\frac{10-x}{x^{2}+10 x+25}\)

Step-by-Step Solution

Answer

\( \frac{10-x}{x^{2}+10 x+25} = \frac{-1}{x+5} + \frac{15}{(x+5)^2} \)

Step 1: Simplify the Quadratic Denominator

The first step is to recognize that the denominator can be factored. Notice that the denominator \(x^2 + 10x + 25\) can be rewritten as \((x+5)^2\).

Step 2: Set Up the Partial Fraction Decomposition

Because the denominator is \((x+5)^2\), the partial fraction decomposition will take the form: \( \frac{A}{x+5} + \frac{B}{(x+5)^2} \).

Step 3: Multiply Through by the Denominator

Multiply the entire equation by \((x+5)^2\) to eliminate the denominators: \[10 - x = A(x+5) + B\]

Step 4: Expand and Simplify the Equation

Expand the right-hand side: \[ A(x+5) + B = Ax + 5A + B \]So the equation becomes: \[ 10 - x = Ax + 5A + B \]

Step 5: Compare Coefficients

Compare coefficients of like terms: 1. For the \(x\) term: \(-1 = A\)2. For the constant term: \(10 = 5A + B\)

Step 6: Solve the Linear System

Substitute \(A = -1\) into the constant term equation:\[10 = 5(-1) + B \10 = -5 + B \B = 15\]

Step 7: Write the Partial Fraction Decomposition

Substitute the values of \(A\) and \(B\) back into the partial fractions: \[ \frac{10-x}{x^{2}+10 x+25} = \frac{-1}{x+5} + \frac{15}{(x+5)^2} \]

Key Concepts

Quadratic Denominator

In partial fraction decomposition, recognizing the form of the denominator is crucial. When dealing with rational expressions, identifying if the denominator is a quadratic expression can greatly simplify your work. For example, the expression \( x^2 + 10x + 25 \) is a quadratic denominator. Quadratics often take the general form \( ax^2 + bx + c \). By inspecting the coefficients, in this case, it becomes apparent that this quadratic can be transformed. The goal is to factor this expression to its simplest form.

In our example, \( x^2 + 10x + 25 \) is a perfect square trinomial. This means it can be rewritten as \((x+5)^2\). Recognizing this allows simplification which sets the stage for partial fraction decomposition.
Here's how it works:

• Identify the quadratic expression in the denominator.
• Look for a pattern like the perfect square trinomial.
• Rewrite accordingly to facilitate easier decomposition.

This strategic simplification aids immensely in the subsequent steps.

Factoring Algebraic Expressions

Factoring algebraic expressions involves breaking down a complex expression into a product of simpler factors. It's an essential skill in algebra, particularly useful in partial fraction decomposition. In scenarios involving quadratic denominators, factoring is often the first step.

The process of factoring requires an understanding of patterns in polynomial expressions, such as difference of squares or perfect square trinomials. For instance, with the expression \(x^2 + 10x + 25\), recognizing it as a perfect square trinomial is key. Here's the step-by-step approach:

• Check if the quadratic is a perfect square, where both the first term \(a^2\) and last term \(b^2\) are squared terms, and the middle term \(2ab\) supports this identity.
• Write it as \((x+5)^2\).

This factorization modifies the original complicated expression, making the decomposition task manageable. Practicing the identification of factorable forms quickly eases the work involved in finding partial fractions.

Rational Expressions

Rational expressions represent a ratio of two polynomials, similar to fractions but in algebraic form. Simplifying and working with rational expressions involves procedures parallel to arithmetic fractions, albeit with more complexity due to algebraic terms.

Understanding the structure of rational expressions is fundamental in algebra. For example, in the expression \(\frac{10-x}{x^2 + 10x + 25}\), it's crucial to simplify wherever possible. This means factoring both the numerator and the denominator, if applicable. Here:

• The numerator \(10-x\) is already in simplest form.
• The denominator \(x^2 + 10x + 25\) is factored into \((x+5)^2\).

This sets the ground for constructing a decomposition that allows the analysis and integration of functions. Simplifying rational expressions simplifies problem-solving and leads to clearer, more manageable expressions for further algebraic work.

Comparing Coefficients

Comparing coefficients is a method used to determine unknown values in polynomial equations by equating coefficients of like terms from both sides of an equation. It is especially useful in partial fraction decomposition to find constants in the numerators of the decomposed fractions.

In our example, after multiplying through by the denominator and simplifying, you reach an equation \(10 - x = Ax + 5A + B\). By comparing corresponding coefficients on both sides, the values of \(A\) and \(B\) are determined:

• For the coefficient of \(x\): \(-1 = A\)
• For the constant term: \(10 = 5A + B\)

Substituting \(A = -1\) into the constant equation solves \(B\). This method is straightforward but requires careful attention to align terms correctly. Practicing with different expressions enhances your ability to solve equations efficiently using this approach.