Question

Decompose \(\left(x^{2}+8\right) /\left(x^{2}-7 x+6\right)\) into partial fractions.

Short answer

The decomposed form of the given function is: \(\frac{-9/5}{x - 1} + \frac{44/5}{x - 6}\).

Step-by-step solution

Factor the denominator

We have to factor the expression in the denominator, which is: \(x^2 - 7x + 6\) This can be factored as: \((x - 1)(x - 6)\) So, our rational function can be written as: \(\frac{x^2 + 8}{(x - 1)(x - 6)}\)

Decompose into partial fractions

Now, let's decompose the fraction into partial fractions. We need to find constants A and B such that: \(\frac{x^2 + 8}{(x - 1)(x - 6)} = \frac{A}{x - 1} + \frac{B}{x - 6}\) To clear the denominators, we multiply both sides of the equation by (x - 1)(x - 6), we get: \(x^2 + 8 = A(x - 6) + B(x - 1)\) Now, we need to solve for A and B.

Solve for A and B

To solve for the constants A and B, we need to substitute different values of x into the equation: \(x^2 + 8 = A(x - 6) + B(x - 1)\) First, let's set x = 1: \(1^2 + 8 = A(1 - 6) + B(1 - 1) \Rightarrow 9 = -5A\) Solving for A, we find that A = -9/5. Now, let's set x = 6: \(6^2 + 8 = A(6 -6) + B(6 -1) \Rightarrow 44 = 5B\) Solving for B, we find that B = 44/5.

Write the decomposed function

Now that we have found the values for A and B, we can rewrite the original fraction as the sum of the partial fractions: \(\frac{x^2 + 8}{(x - 1)(x - 6)} = \frac{-9/5}{x - 1} + \frac{44/5}{x - 6}\) So, the decomposed form of the given function is: \(\frac{-9/5}{x - 1} + \frac{44/5}{x - 6}\)

Key concepts

Factoring Quadratics

Factoring quadratics is a key step in simplifying many algebraic expressions, including partial fraction decomposition. When factoring a quadratic expression like

• \(x^2 - 7x + 6\),

start by identifying two numbers that multiply to give you the constant term (6 in this case), and add up to the coefficient of the linear term (-7). These numbers are -1 and -6 because

• \((-1) \times (-6) = 6\)
• \((-1) + (-6) = -7\).

This means, you can rewrite the quadratic as

• \((x - 1)(x - 6)\),

changing your expression to make it easier to work with.
Factoring quadratics can often involve trial and error, especially if numbers aren't immediately obvious. It becomes a lot easier if you're familiar with common factorizations or formulas for more stubborn cases.

Rational Functions

Rational functions are a type of function defined as the ratio of two polynomials. For example, the given expression

• \(\frac{x^2 + 8}{x^2 - 7x + 6}\) is a rational function.

Rational functions have asymptotes, which are lines the graph of the function approaches but never touches. The roots of the denominator (values that make the denominator zero) often determine vertical asymptotes.
The role of rational functions in mathematics is important as they can express many real-world phenomena and help in computational problems, making them common in calculus and algebra explorations.

Solving Linear Equations

Linear equations often appear when finding constants in partial fraction decomposition. To solve them, we substitute specific values for variables to simplify the process. In our example, we set

• x to 1 and 6,

so the terms involving \(x\) drop out and simplify the calculations.

• For \(x = 1\), the equation becomes \(9 = -5A\), solving gives \(A = -9/5\).
• For \(x = 6\), simplifying gives \(44 = 5B\), leading to \(B = 44/5\).

This systematic approach using specific values makes solving for constants straightforward. Whatever the method, solving these equations requires balancing or canceling terms systematically to isolate the unknown.

Constants in Equations

Constants play a vital role in solving algebraic equations, especially in partial fraction decomposition. They allow us to express complex expressions as simpler components. After rearranging terms and substituting values, we use constants

• (like A and B in our example)

to express parts of the equation individually and simplify our rational function. These constants are crucial for translating a complex rational function into partial fractions that are easier to analyze mathematically.
To find the constants correctly, consistency in substitution and calculation is crucial. Mistakes in establishing these values can skew your resulting decomposition, preventing accurate representation of the initial function.