Expressing (2x^2 + X - 3) / ((x^2 - 2)(x + 1)) In Partial Fractions

Introduction

In the realm of mathematics, particularly in algebra and calculus, the technique of partial fraction decomposition plays a pivotal role. Partial fraction decomposition is a powerful method used to break down complex rational expressions into simpler fractions. This technique is invaluable when integrating rational functions, solving differential equations, or performing other algebraic manipulations. In this comprehensive guide, we will delve deep into the process of expressing the rational expression $(2x^2 + x - 3) / ((x^2 - 2)(x + 1))$ in partial fractions. Guys, this is a classic problem, and mastering it will seriously level up your math skills!

Understanding Partial Fractions

Before we dive into the specifics of our problem, let's first grasp the fundamental concept of partial fractions. A rational expression is simply a fraction where both the numerator and the denominator are polynomials. The goal of partial fraction decomposition is to rewrite a complicated rational expression as a sum of simpler fractions, each with a simpler denominator.

Think of it like this: imagine you have a cake that's been sliced into different sized pieces. Partial fraction decomposition is like figuring out how to put those pieces back together to make the whole cake, but instead of cake, we're dealing with algebraic fractions. The main reason we do this is that these simpler fractions are often much easier to work with, especially when we need to integrate them or perform other calculus operations. Trust me, your future calculus self will thank you for understanding this!

Setting Up the Partial Fraction Decomposition

Now, let's tackle the problem at hand: expressing $(2x^2 + x - 3) / ((x^2 - 2)(x + 1))$ in partial fractions. The first step is to recognize the structure of the denominator. We have $(x^2 - 2)(x + 1)$, which means we have a quadratic factor $(x^2 - 2)$ and a linear factor $(x + 1)$. This tells us how to set up the partial fraction decomposition.

For the quadratic factor $(x^2 - 2)$, we need a linear expression in the numerator, which we'll represent as $Ax + B$. For the linear factor $(x + 1)$, we simply need a constant in the numerator, which we'll call $C$. So, our partial fraction decomposition will look like this:

$$\frac{2x^2 + x - 3}{(x^2 - 2)(x + 1)} = \frac{Ax + B}{x^2 - 2} + \frac{C}{x + 1}$$

The next step is crucial: we need to find the values of the constants $A$, $B$, and $C$. This is where the real algebraic fun begins!

Solving for the Constants

To find the values of $A$, $B$, and $C$, we need to clear the denominators. We do this by multiplying both sides of the equation by the original denominator, which is $(x^2 - 2)(x + 1)$. This gives us:

$$2x^2 + x - 3 = (Ax + B)(x + 1) + C(x^2 - 2)$$

Now, we expand the right side of the equation:

$$2x^2 + x - 3 = Ax^2 + Ax + Bx + B + Cx^2 - 2C$$

Next, we group like terms:

$$2x^2 + x - 3 = (A + C)x^2 + (A + B)x + (B - 2C)$$

Now we have a polynomial equation, and for this equation to hold true for all values of $x$, the coefficients of the corresponding terms must be equal. This gives us a system of linear equations:

• Coefficient of $x^2$: $A + C = 2$
• Coefficient of $x$: $A + B = 1$
• Constant term: $B - 2C = -3$

We now have a system of three equations with three unknowns. There are several ways to solve this system, such as substitution, elimination, or matrix methods. Let's use substitution.

From the first equation, we can express $A$ in terms of $C$: $A = 2 - C$. From the second equation, we can express $B$ in terms of $A$: $B = 1 - A$.

Substitute $A = 2 - C$ into the equation for $B$: $B = 1 - (2 - C) = C - 1$.

Now substitute $B = C - 1$ into the third equation: $(C - 1) - 2C = -3$.

Simplify and solve for $C$:

$$-C - 1 = -3$$

$$-C = -2$$

$$C = 2$$

Now that we have $C$, we can find $A$ and $B$:

$$A = 2 - C = 2 - 2 = 0$$

$$B = C - 1 = 2 - 1 = 1$$

So, we have found that $A = 0$, $B = 1$, and $C = 2$. These are the constants we need for our partial fraction decomposition.

Writing the Partial Fraction Decomposition

Now that we have the values of $A$, $B$, and $C$, we can write out the partial fraction decomposition:

$$\frac{2x^2 + x - 3}{(x^2 - 2)(x + 1)} = \frac{0x + 1}{x^2 - 2} + \frac{2}{x + 1}$$

Simplifying, we get:

$$\frac{2x^2 + x - 3}{(x^2 - 2)(x + 1)} = \frac{1}{x^2 - 2} + \frac{2}{x + 1}$$

And there you have it! We have successfully expressed the given rational expression in partial fractions. This form is much easier to work with in many situations, such as integration.

Conclusion

In this guide, we've walked through the process of expressing the rational expression $(2x^2 + x - 3) / ((x^2 - 2)(x + 1))$ in partial fractions. We started by understanding the concept of partial fraction decomposition and its importance. Then, we set up the decomposition, solved for the constants, and wrote out the final result.

Partial fraction decomposition is a powerful tool in mathematics, particularly in calculus and algebra. By mastering this technique, you'll be well-equipped to tackle a wide range of problems involving rational functions. So keep practicing, and you'll become a partial fraction pro in no time! Remember, guys, math is all about practice, so keep at it!

Further Practice

To solidify your understanding, try applying this technique to other rational expressions. You can create your own problems or find examples in textbooks or online resources. The more you practice, the more comfortable you'll become with the process. Good luck, and happy decomposing!

Key Takeaways

• Partial fraction decomposition breaks down complex rational expressions into simpler fractions.
• The structure of the denominator determines the form of the partial fractions.
• Solving for the constants involves clearing denominators and solving a system of equations.
• The resulting partial fractions are often easier to work with than the original expression.

So, that’s it for this guide on partial fractions! I hope this has been helpful and that you now feel more confident in your ability to tackle these types of problems. Keep practicing, and you’ll become a math wizard in no time. Remember, guys, math is like a muscle – the more you use it, the stronger it gets!

If you have any questions or want to share your own experiences with partial fractions, feel free to leave a comment below. And don’t forget to check out other math resources to continue your learning journey. Happy mathing!

Practice Problems

To really nail this down, let's try a few more examples. Working through these will help solidify your understanding and make you a partial fraction master.

Example 1: A Simpler Case

Let’s start with something a little less complex. Express the following rational expression in partial fractions:

$$\frac{5x - 4}{x^2 - x - 2}$$

First, we factor the denominator:

$$x^2 - x - 2 = (x - 2)(x + 1)$$

Now, we set up the partial fraction decomposition:

$$\frac{5x - 4}{(x - 2)(x + 1)} = \frac{A}{x - 2} + \frac{B}{x + 1}$$

Multiply both sides by the denominator $(x - 2)(x + 1)$ to clear fractions:

$$5x - 4 = A(x + 1) + B(x - 2)$$

Expand and group like terms:

$$5x - 4 = Ax + A + Bx - 2B$$

$$5x - 4 = (A + B)x + (A - 2B)$$

Now, we set up the system of equations:

• $A + B = 5$
• $A - 2B = -4$

We can solve this system using substitution or elimination. Let's use substitution. From the first equation, we have $A = 5 - B$. Substitute this into the second equation:

$$(5 - B) - 2B = -4$$

$$5 - 3B = -4$$

$$-3B = -9$$

$$B = 3$$

Now, find $A$:

$$A = 5 - B = 5 - 3 = 2$$

So, $A = 2$ and $B = 3$. Now, we can write the partial fraction decomposition:

$$\frac{5x - 4}{x^2 - x - 2} = \frac{2}{x - 2} + \frac{3}{x + 1}$$

Example 2: Repeated Linear Factors

Now, let's try a case with repeated linear factors. Express the following in partial fractions:

$$\frac{x + 2}{(x - 1)^2}$$

Since we have a repeated factor $(x - 1)^2$, the partial fraction decomposition looks like this:

$$\frac{x + 2}{(x - 1)^2} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2}$$

Multiply both sides by $(x - 1)^2$:

$$x + 2 = A(x - 1) + B$$

Expand:

$$x + 2 = Ax - A + B$$

Now, equate coefficients:

• Coefficient of $x$: $A = 1$
• Constant term: $-A + B = 2$

Since $A = 1$, we can find $B$:

$$-1 + B = 2$$

$$B = 3$$

So, the partial fraction decomposition is:

$$\frac{x + 2}{(x - 1)^2} = \frac{1}{x - 1} + \frac{3}{(x - 1)^2}$$

Example 3: Irreducible Quadratic Factors

Let's tackle one more example with an irreducible quadratic factor. Express the following in partial fractions:

$$\frac{3x^2 + 5x - 2}{(x^2 + 1)(x - 1)}$$

For the irreducible quadratic factor $x^2 + 1$, we need a linear term in the numerator. So, the partial fraction decomposition is:

$$\frac{3x^2 + 5x - 2}{(x^2 + 1)(x - 1)} = \frac{Ax + B}{x^2 + 1} + \frac{C}{x - 1}$$

Multiply both sides by $(x^2 + 1)(x - 1)$:

$$3x^2 + 5x - 2 = (Ax + B)(x - 1) + C(x^2 + 1)$$

Expand:

$$3x^2 + 5x - 2 = Ax^2 - Ax + Bx - B + Cx^2 + C$$

Group like terms:

$$3x^2 + 5x - 2 = (A + C)x^2 + (-A + B)x + (-B + C)$$

Equate coefficients:

• $A + C = 3$
• $-A + B = 5$
• $-B + C = -2$

Let's solve this system. From the first equation, $C = 3 - A$. From the second equation, $B = 5 + A$. Substitute these into the third equation:

$$-(5 + A) + (3 - A) = -2$$

$$-5 - A + 3 - A = -2$$

$$-2A - 2 = -2$$

$$-2A = 0$$

$$A = 0$$

Now, find $B$ and $C$:

$$B = 5 + A = 5 + 0 = 5$$

$$C = 3 - A = 3 - 0 = 3$$

So, the partial fraction decomposition is:

$$\frac{3x^2 + 5x - 2}{(x^2 + 1)(x - 1)} = \frac{0x + 5}{x^2 + 1} + \frac{3}{x - 1}$$

Simplify:

$$\frac{3x^2 + 5x - 2}{(x^2 + 1)(x - 1)} = \frac{5}{x^2 + 1} + \frac{3}{x - 1}$$

These examples should give you a good feel for how to handle different types of partial fraction decomposition problems. Remember, the key is to practice, practice, practice! The more you work through these, the more comfortable and confident you’ll become. Keep up the great work, guys!

Now, go forth and conquer those fractions!