Question

Resolve \(\frac{1}{x^{2}-7 x-18}\) into partial fractions. (1) \(\frac{1}{11}\left(\frac{1}{x-9}-\frac{1}{x+2}\right)\) (2) \(\frac{1}{11}\left(\frac{1}{x-9}+\frac{1}{x+2}\right)\) (3) \(\frac{1}{11}\left(\frac{1}{x+9}-\frac{1}{x-2}\right)\) (4) \(\frac{1}{11}\left(\frac{1}{x+9}+\frac{1}{x+2}\right)\)

Short answer

(1) \(\frac{1}{11}\left(\frac{1}{x-9}-\frac{1}{x+2}\right)\)

Step-by-step solution

Factor the Denominator

To begin, first, factor the denominator of the given rational function. In this case, we have the polynomial: \(x^2 - 7x - 18\) This will factor into the product of two linear terms: \((x - 9)(x + 2)\).

Set Up Partial Fractions

Now that we have factored the denominator, we can set up the partial fractions. Let: \(\frac{1}{(x - 9)(x + 2)} = \frac{A}{x - 9} + \frac{B}{x + 2}\)

Clear the Denominator

To find A and B, we will clear the denominator by multiplying both sides by \((x - 9)(x + 2)\): \(1 = A(x + 2) + B(x - 9)\)

Solve for A and B

To find the values of A and B, we can use substitution. We will substitute the values where the denominators of the partial fractions are zero. For A: Let \(x = 9\), so \(1 = A(11) + B(0)\), which gives \(A = \frac{1}{11}\). For B: Let \(x = -2\), so \(1 = A(0) + B(-11)\), which gives \(B = -\frac{1}{11}\).

Write the Partial Fraction Decomposition

Now that we have the values of A and B, we can write the partial fraction decomposition of the given function: \(\frac{1}{x^2 - 7x - 18} = \frac{1}{11} \left( \frac{1}{x - 9} - \frac{1}{x + 2} \right)\) The correct answer is (1) \(\frac{1}{11}\left(\frac{1}{x-9}-\frac{1}{x+2}\right)\).

Key concepts

Rational Functions

Rational functions are expressions that form fractions where the numerator and the denominator are both polynomials.
This means that in any rational function, you'll see a polynomial on top and another one on the bottom of the fraction.
To understand rational functions, think about dividing one polynomial by another.
This division gives us a fraction that behaves differently depending on how complex each polynomial is.
Usually, we're interested in cases where the degree of the polynomial in the denominator is equal to or greater than the one in the numerator. In terms of partial fractions, we take these rational functions and split them into simpler fractions.
These simpler fractions are usually much easier to work with, especially in solving calculus problems.

Factorization

Factorization is a method of breaking down an equation into simpler components or 'factors' that multiply together to give the original equation.
Consider, for example, the quadratic expression from our exercise: \(x^2 - 7x - 18\).
Factorization involves finding two numbers that multiply to the constant term, in this case, -18, and add to the linear coefficient, which is -7.
For our exercise, these numbers are -9 and +2.
Therefore, \(x^2 - 7x - 18\) can be factored into the linear terms \((x - 9)\) and \((x + 2)\).
By converting to these linear terms, we simplify the problem of resolving into partial fractions.

Denominator

The denominator is the bottom part of a fraction.
It tells us the 'whole' into which our numerator is divided.
In the context of rational functions and partial fractions, the denominator is particularly important.To successfully decompose a fraction into partial fractions, the denominator must be factorized first.
In our solved example, the denominator was \(x^2 - 7x - 18\).
Upon factorization, we find it can be expressed as \((x - 9)(x + 2)\).
Once in this form, the denominator allows us to break the fraction into parts that are easier to analyze and solve.
This showcases why understanding and working with the denominator is crucial in mathematical manipulation.

Linear Terms

Linear terms are expressions where the highest power of the variable is one.
They are considered simple because they are the simplest form of polynomial.In partial fractions, we often break down complex expressions using these linear terms.
In the example given, the rational function's denominator \((x^2 - 7x - 18)\) factors into two linear terms: \((x - 9)\) and \((x + 2)\).
Linear terms make it simpler to resolve the function to partial fractions because they convert it into a sum of easily manageable pieces.
With linear terms, we reach equations that are linear, which are straightforward to solve compared to those involving higher powers.
Using these terms effectively can simplify complex algebraic problems, especially in calculus or differential equations.