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Write the form of the partial fraction decomposition of the rational expression. It is not necessary to solve for the constants. $$\frac{6 x^{2}-14 x-27}{(x+2)(x-3)^{2}}$$

Short Answer

Expert verified
\(\frac {6 x^{2}-14 x-27} {(x+2)(x-3)^{2}} = \frac{A}{x+2} + \frac{B}{x - 3} + \frac {C}{(x - 3)^2}\)

Step by step solution

01

Recognize the Structure

Looking at the denominator of the given rational expression, the factors are \( (x+2) \) and \( (x-3)^2 \). This form suggests that when the rational expression is broken down, each unique factor from the denominator (i.e., \( x+2 \), \( x-3 \)) will correspond to a term in the decomposition.
02

Assign Appropriate Forms

We know that a linear term in the denominator like \( (x + a) \) corresponds to \(\frac {A} {x + a}\) in the decomposition. Furthermore, a repeated linear term like \( (x + a)^n \) corresponds to \(\frac {A_1} {x + a} + \frac {A_2} {x + a}^2 + \ldots + \frac {A_n} {x + a}^n\). Hence, applying this to our expression, the term corresponding to \( x + 2 \) is \(\frac {A} {x + 2}\) and the term corresponding to \( (x - 3)^2 \) is \(\frac {B} {x - 3} + \frac {C} {(x - 3)^2}\).
03

Write the Decomposition

Combine the forms assigned in the previous step to obtain the full decomposition of the expression. This results into \(\frac {6 x^{2}-14 x-27} {(x+2) (x-3)^{2}} = \frac{A}{x+2} + \frac{B}{x - 3} + \frac{C}{(x - 3)^2}\). We have not been asked to solve for the constants \(A\), \(B\), and \(C\), therefore this is our final result.

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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 akin to fractions but with polynomials in both the numerator and the denominator. Just like fractions, they can be simplified, added, subtracted, multiplied and divided. The key to mastering rational expressions is to understand the relationship between the numerator and the denominator.

A crucial skill in handling rational expressions is recognizing when and how to apply partial fraction decomposition, especially when integrating or solving equations. This process transforms a complex rational expression into a sum of simpler fractions, whose denominators are the factorized components of the original denominator.

Our example expression, \(\frac{6 x^{2}-14 x-27}{(x+2)(x-3)^{2}}\), demonstrates this perfectly. By factoring the denominator, you can work towards breaking down the expression into terms that are more manageable, providing a cornerstone for integral calculus and advanced algebra.
Algebraic Fractions
Algebraic fractions, or polynomial fractions, are the division of two polynomials. Just like the fractions you've learned with numbers, algebraic fractions follow similar rules but involve variables. Simplification, one of the essential skills, helps reduce the fraction to its lowest terms by canceling out common factors from the numerator and the denominator.

In practice, such as with the question at hand \(\frac{6 x^{2}-14 x-27}{(x+2)(x-3)^{2}}\), we can also be required to split the fraction into parts to simplify integration or to solve equations. Each part of the fraction after decomposition corresponds to a factor in the denominator, assisting in isolating particular components of the expression.
Factoring Polynomials
Factoring is a method of breaking down polynomials into products of simpler polynomials that, when multiplied together, give back the original polynomial. Mastering factoring techniques, such as factoring out the greatest common factor, using difference of squares, or applying the factor theorem, is fundamental for simplifying algebraic fractions and solving polynomial equations.

Our provided example involves a denominator that factored into \( (x+2)(x-3)^{2} \). Recognizing the square of the binomial \( (x-3)^{2} \) is particularly important since each occurrence of the factor \( (x-3) \) will correspond to a different term in the partial fraction decomposition. The ability to spot and factor such expressions is a critical step in various areas of algebra, precalculus, and calculus.

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