Partial Fractions: Express (33-9x)/(x^2-6x+9)

Hey guys! Today, we're diving into a fun little problem from the world of mathematics: expressing the fraction (33 - 9x) / (x^2 - 6x + 9) in partial fractions. This might sound intimidating at first, but trust me, it's a super useful technique, especially when you're dealing with integrals or inverse Laplace transforms in calculus and engineering. So, let's break it down step by step and make sure we all get it.

Understanding Partial Fractions

Before we jump into the specifics of our problem, let's quickly recap what partial fractions are all about. In essence, partial fraction decomposition is a method that allows us to break down a complex rational expression (a fraction where the numerator and denominator are polynomials) into simpler fractions. Think of it like reverse engineering how fractions are added together. Instead of combining simple fractions into a more complex one, we're doing the opposite – taking a complex fraction and splitting it into its simpler building blocks. This is incredibly handy because these simpler fractions are often much easier to work with in various mathematical operations. The key idea here is to rewrite the given rational function as a sum of simpler fractions, each with a denominator that is a factor of the original denominator. This makes the function much easier to integrate, differentiate, or otherwise manipulate.

The main goal of partial fraction decomposition is to simplify complex rational expressions. We start with a rational function, which is basically a fraction where both the numerator and the denominator are polynomials. Imagine trying to integrate a complicated fraction directly – it can be a nightmare! But, if we can break it down into simpler fractions, each one is much easier to handle. This is why partial fractions are so useful in calculus, especially when dealing with integration problems. In engineering, you'll often encounter situations where you need to analyze systems using transfer functions, which are often rational expressions. Partial fraction decomposition helps in finding the inverse Laplace transform, which is crucial for understanding the time-domain behavior of the system. For example, when analyzing circuits or control systems, you might need to determine how a system responds to a particular input over time. By breaking down the transfer function into partial fractions, you can easily find the inverse Laplace transform and get a clear picture of the system's response. Partial fractions also pop up in other areas of math, like solving differential equations. Sometimes, you'll encounter a differential equation that's hard to solve in its original form. But, by using techniques like Laplace transforms and partial fractions, you can convert the differential equation into an algebraic equation, solve it, and then transform the solution back to the original domain. This can make a seemingly impossible problem much more manageable. So, understanding partial fractions isn't just about manipulating fractions; it's a powerful tool that unlocks solutions in various fields. Whether you're dealing with integrals, system analysis, or differential equations, this technique can be a real game-changer.

Factoring the Denominator

Alright, let's get our hands dirty with the problem at hand. The first crucial step in expressing (33 - 9x) / (x^2 - 6x + 9) in partial fractions is to factor the denominator, which is x^2 - 6x + 9. Factoring is like finding the puzzle pieces that, when multiplied together, give you the original expression. In this case, we're looking for two binomials that multiply to x^2 - 6x + 9. If you're familiar with factoring quadratic expressions, you might already recognize this as a perfect square trinomial. But if not, no worries, we'll walk through it.

We need to find two numbers that add up to -6 (the coefficient of the x term) and multiply to 9 (the constant term). Those numbers are -3 and -3. So, we can factor the denominator as (x - 3)(x - 3), which can also be written as (x - 3)^2. This is an important observation because it tells us that we have a repeated linear factor in the denominator. Repeated factors require a slightly different approach when we set up our partial fractions, so keep this in mind. Factoring the denominator is the cornerstone of partial fraction decomposition. It’s the first step that sets the stage for the rest of the process. When you factor the denominator, you're essentially identifying the building blocks of the original fraction. Each factor in the denominator will correspond to a partial fraction in the decomposition. For instance, if the denominator factors into (x - a)(x - b), you'll have two partial fractions, one with a denominator of (x - a) and the other with a denominator of (x - b). The type of factors you encounter – whether they are linear, quadratic, or repeated – will determine the form of the partial fractions. Linear factors like (x - a) are the simplest, leading to partial fractions of the form A / (x - a). Quadratic factors that can't be factored further require a bit more attention, often resulting in partial fractions of the form (Ax + B) / (x^2 + bx + c). And, as we've seen in our example, repeated factors like (x - 3)^2 need special treatment, which we'll dive into shortly. So, mastering the art of factoring is absolutely crucial for tackling partial fraction problems. It’s like having the right key to unlock the solution. Make sure you’re comfortable with various factoring techniques, from simple factoring of quadratics to recognizing patterns like the difference of squares or perfect square trinomials. The more proficient you are at factoring, the smoother the rest of the partial fraction decomposition process will be. Trust me, spending the time to sharpen your factoring skills will pay off big time when you're faced with complex rational expressions. It's the foundation upon which everything else is built.

Setting Up the Partial Fractions

Now that we've factored the denominator as (x - 3)^2, we know we're dealing with a repeated linear factor. This means our partial fraction setup will look a little different than if we had distinct linear factors. For a repeated factor like (x - 3)^2, we need to include a term for each power of the factor, up to the highest power. So, in this case, we'll have two terms: one with a denominator of (x - 3) and another with a denominator of (x - 3)^2. Our partial fraction decomposition will look like this:

(33 - 9x) / (x - 3)^2 = A / (x - 3) + B / (x - 3)^2

Where A and B are constants that we need to determine. This setup is crucial for correctly decomposing the fraction. If you miss a term or don't account for the repeated factor properly, you won't be able to find the correct values for A and B. Setting up the partial fractions correctly is a critical step because it lays the foundation for solving for the unknown constants. The way you set up the partial fractions depends entirely on the factors you found in the denominator. As we discussed earlier, different types of factors require different approaches. For each distinct linear factor (x - a), you'll have a partial fraction of the form A / (x - a), where A is a constant you need to find. If you have a repeated linear factor like (x - a)^n, you'll need to include n partial fractions, one for each power of the factor, like A1 / (x - a) + A2 / (x - a)^2 + ... + An / (x - a)^n. This is what we did in our example with the (x - 3)^2 factor. For each irreducible quadratic factor (ax^2 + bx + c), which can't be factored further using real numbers, you'll have a partial fraction of the form (Ax + B) / (ax^2 + bx + c), where A and B are constants to be determined. The key is to make sure you have a partial fraction term for every factor in the denominator, and that you account for repeated factors correctly. If you skip a term or set up the fractions incorrectly, you'll end up with the wrong system of equations, and you won't be able to solve for the constants accurately. So, take your time, double-check your factors, and make sure your partial fraction setup matches the structure of the denominator. This will save you a lot of headaches down the road. The process is systematic, but it requires careful attention to detail. Think of it like building a house – the foundation has to be solid before you can start adding the walls and the roof. A correct setup is the solid foundation for your partial fraction decomposition.

Solving for the Constants

Now comes the fun part – solving for the constants A and B! To do this, we'll start by multiplying both sides of our equation by the original denominator, which is (x - 3)^2. This will clear the fractions and leave us with a polynomial equation. So, let's do it:

(33 - 9x) / (x - 3)^2 = A / (x - 3) + B / (x - 3)^2

Multiply both sides by (x - 3)^2:

33 - 9x = A(x - 3) + B

Now, we have a couple of ways to solve for A and B. One method is to expand the right side and equate coefficients. This means we'll group the terms with the same powers of x and set their coefficients equal to each other. The other method, which is often quicker in this case, is to substitute convenient values of x that will eliminate one of the variables. Let's use the substitution method first. Notice that if we let x = 3, the term with A will become zero, leaving us with an equation we can easily solve for B. So, let's try that:

Let x = 3:

33 - 9(3) = A(3 - 3) + B

33 - 27 = B

6 = B

Great! We've found that B = 6. Now, we need to find A. Since we've already used x = 3, we can substitute any other value for x. A simple choice is x = 0:

Let x = 0:

33 - 9(0) = A(0 - 3) + 6

33 = -3A + 6

27 = -3A

A = -9

So, we've found that A = -9 and B = 6. We have successfully solved for our constants! Solving for the constants is a critical step, and there are several techniques you can use. The method of equating coefficients is a classic approach. After clearing the fractions, you expand the equation and group terms with the same powers of x. Then, you set the coefficients of corresponding terms on both sides of the equation equal to each other. This gives you a system of linear equations that you can solve for the constants. For example, if you have an equation like 5x + 3 = Ax + B, you can equate the coefficients of x to get A = 5 and equate the constant terms to get B = 3. This method is very systematic and works well for any partial fraction decomposition problem. The substitution method, as we used in our example, involves choosing strategic values of x that simplify the equation and allow you to solve for the constants more easily. This often works well when you have linear factors in the denominator. If you have a factor like (x - a), substituting x = a will make the term with that factor equal to zero, which can help you isolate the other constants. The key is to choose values of x that will eliminate some of the unknowns, making it easier to solve for the remaining ones. Sometimes, a combination of both methods is the most efficient way to solve for the constants. You might use substitution to find one or two constants and then use equating coefficients to find the rest. The best approach depends on the specific problem and your personal preference. The important thing is to be comfortable with both methods and to be able to choose the one that will work best in a given situation. Solving for the constants is like cracking a code – you're using algebraic techniques to unlock the values that make the partial fraction decomposition work. Once you've found the constants, you're one step closer to expressing the original fraction in its simpler form.

Writing the Partial Fractions

Now that we've found A = -9 and B = 6, we can write out the partial fraction decomposition. We simply substitute these values back into our setup:

(33 - 9x) / (x^2 - 6x + 9) = -9 / (x - 3) + 6 / (x - 3)^2

And there you have it! We've successfully expressed the original fraction in partial fractions. This is our final answer. Writing the partial fractions is the final step in the process, and it's where everything comes together. After you've solved for the constants, you simply plug them back into the partial fraction setup you created earlier. This gives you the decomposition of the original rational expression into simpler fractions. It's a bit like putting the pieces of a puzzle together – you've found all the individual components, and now you're assembling them to see the complete picture. In our example, we found that A = -9 and B = 6, so we substituted these values into our setup: (33 - 9x) / (x^2 - 6x + 9) = -9 / (x - 3) + 6 / (x - 3)^2. This final expression shows the original fraction broken down into two simpler fractions, each with a denominator that is a factor of the original denominator. This is the power of partial fraction decomposition – it allows you to take a complex rational expression and rewrite it in a form that is often much easier to work with. Whether you're integrating, finding inverse Laplace transforms, or solving differential equations, these simpler fractions can make your life much easier. So, writing out the partial fractions isn't just a formality; it's the culmination of all your hard work. It's the moment where you see the original fraction transformed into its simpler components, ready to be used in further calculations or analysis. Make sure you double-check your work at this stage, ensuring that you've substituted the correct values for the constants and that your final expression matches the form you expected. With a little practice, you'll become a pro at writing partial fractions and using them to solve a wide range of mathematical problems.

Conclusion

So, guys, we've successfully expressed (33 - 9x) / (x^2 - 6x + 9) in partial fractions. We factored the denominator, set up the partial fractions, solved for the constants, and wrote out our final answer. Remember, partial fraction decomposition is a powerful tool that can simplify complex rational expressions and make them easier to work with in various mathematical contexts. Keep practicing, and you'll master this technique in no time! Expressing rational functions in partial fractions is a fundamental technique with far-reaching applications. It's not just a mathematical exercise; it's a skill that opens doors to solving problems in calculus, differential equations, engineering, and beyond. The ability to break down complex fractions into simpler components is like having a secret weapon in your mathematical arsenal. It allows you to tackle problems that might seem daunting at first glance. Whether you're integrating a rational function, finding the inverse Laplace transform of a transfer function, or solving a system of differential equations, partial fraction decomposition can be the key to unlocking the solution. The process we've walked through – factoring the denominator, setting up the partial fractions, solving for the constants, and writing the final decomposition – is a systematic approach that you can apply to any rational function. It requires attention to detail and a solid understanding of algebraic techniques, but with practice, it becomes second nature. Remember, the more you practice, the more comfortable you'll become with the different types of factors you might encounter – linear, quadratic, repeated – and the appropriate setup for each. And don't be afraid to use different methods for solving for the constants, like equating coefficients or strategic substitution. The best approach often depends on the specific problem and your personal preference. So, keep honing your skills, keep exploring the applications of partial fractions, and you'll find yourself well-equipped to tackle a wide range of mathematical challenges. The journey of learning mathematics is a journey of building tools and techniques, and partial fraction decomposition is definitely one of the most valuable tools you can add to your toolbox. Keep practicing, keep exploring, and keep pushing your mathematical boundaries! Well done, guys!