Find Actual Roots Using Rational Root Theorem

Hey everyone! Today, we're diving deep into the fascinating world of polynomials, and specifically, how to find their roots. You know, those magical numbers that make a polynomial equal to zero? We're going to tackle a problem involving the Rational Root Theorem, a super handy tool that helps us narrow down the possibilities. The polynomial we're working with is $f(x)=10 x^3+29 x^2-66 x+27$. We've been given a list of potential roots, and our mission, should we choose to accept it, is to figure out which of these are the actual roots. Get ready to flex those math muscles, guys!

Understanding the Rational Root Theorem

So, what exactly is this Rational Root Theorem, you ask? Think of it as a guidebook for finding rational roots. A rational root is simply a root that can be expressed as a fraction $p/q$, where $p$ and $q$ are integers, and $q$ is not zero. The theorem states that if a polynomial has integer coefficients (which ours does!), then any rational root must be of the form $p/q$, where $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient. This is a huge deal because it gives us a finite list of candidates to test, instead of guessing wildly! For our polynomial, $f(x)=10 x^3+29 x^2-66 x+27$, the constant term is 27 and the leading coefficient is 10.

Let's break down the factors. The factors of the constant term (27) are $\pm 1, \pm 3, \pm 9, \pm 27$. These are our potential values for $p$. The factors of the leading coefficient (10) are $\pm 1, \pm 2, \pm 5, \pm 10$. These are our potential values for $q$. Now, to find all possible rational roots ($p/q$), we need to combine these. We'll take each factor of 27 and divide it by each factor of 10. This will generate a list of all possible rational roots. For example, we could have $1/1$, $1/2$, $1/5$, $1/10$, $3/1$, $3/2$, $3/5$, $3/10$, and so on. It might seem like a lot, but it's way better than an infinite number of possibilities! The Rational Root Theorem is a powerful mathematical principle that simplifies the search for rational roots by providing a systematic method based on the polynomial's coefficients. It's like having a map to navigate the complex landscape of polynomial equations. When you're dealing with polynomials, especially higher-degree ones, finding the roots can feel like searching for a needle in a haystack. The Rational Root Theorem comes to the rescue by giving you a set of specific, finite candidates to test. This dramatically reduces the amount of trial and error involved. Remember, this theorem only guarantees that if there are rational roots, they will be in the list generated by $p/q$. It doesn't tell you if there are any rational roots, nor does it help find irrational or complex roots. But for problems where you're given potential rational roots, like in our case, it's the key to unlocking the solution. The theorem is built on the fundamental properties of polynomial equations and the nature of rational numbers. By looking at the factors of the constant term and the leading coefficient, we are essentially leveraging the structure of the polynomial itself to guide our search. It's a testament to how deeply interconnected mathematical concepts are.

The Candidates: A Closer Look

We've been given a specific list of potential roots: A. $-\frac{9}{2}$, B. $-\frac{9}{10}$, C. $\frac{3}{5}$, D. 1, E. 3. Now, the Rational Root Theorem helped us understand how these potential roots could arise from the polynomial's coefficients. For instance, $-\frac{9}{2}$ fits the $p/q$ form, where $p = -9$ (a factor of 27) and $q = 2$ (a factor of 10). Similarly, $-\frac{9}{10}$ has $p = -9$ and $q = 10$. $\frac{3}{5}$ has $p = 3$ (a factor of 27) and $q = 5$ (a factor of 10). The number 1 can be written as $1/1$, where $p = 1$ and $q = 1$. And 3 can be written as $3/1$, where $p = 3$ and $q = 1$. So, all these options are indeed potential rational roots according to the theorem. Our task now is to test each one to see if it actually makes $f(x)$ equal to zero.

Testing the Potential Roots

Alright, it's time for the main event: plugging these numbers into our polynomial $f(x)=10 x^3+29 x^2-66 x+27$ and seeing if we get 0. This is where the real work happens. We’ll go through each option one by one. If $f(c) = 0$ for a candidate $c$, then $c$ is an actual root!

Testing Option A: $-\frac{9}{2}$

Let's substitute $x = -9/2$ into $f(x)$: $f(-\frac{9}{2}) = 10(-\frac{9}{2})^3 + 29(-\frac{9}{2})^2 - 66(-\frac{9}{2}) + 27$ $f(-\frac{9}{2}) = 10(-\frac{729}{8}) + 29(\frac{81}{4}) - 66(-\frac{9}{2}) + 27$ $f(-\frac{9}{2}) = -\frac{7290}{8} + \frac{2349}{4} + \frac{594}{2} + 27$ To add these, we need a common denominator, which is 8: $f(-\frac{9}{2}) = -\frac{7290}{8} + \frac{2349 \times 2}{8} + \frac{594 \times 4}{8} + \frac{27 \times 8}{8}$ $f(-\frac{9}{2}) = -\frac{7290}{8} + \frac{4698}{8} + \frac{2376}{8} + \frac{216}{8}$ $f(-\frac{9}{2}) = \frac{-7290 + 4698 + 2376 + 216}{8}$ $f(-\frac{9}{2}) = \frac{-7290 + 7290}{8}$ $f(-\frac{9}{2}) = \frac{0}{8} = 0$

Wowza! Option A, $-\frac{9}{2}$, is indeed an actual root because it makes the polynomial equal to zero. That's one down, four to go!

Testing Option B: $-\frac{9}{10}$

Now, let's try $x = -9/10$: $f(-\frac{9}{10}) = 10(-\frac{9}{10})^3 + 29(-\frac{9}{10})^2 - 66(-\frac{9}{10}) + 27$ $f(-\frac{9}{10}) = 10(-\frac{729}{1000}) + 29(\frac{81}{100}) - 66(-\frac{9}{10}) + 27$ $f(-\frac{9}{10}) = -\frac{7290}{1000} + \frac{2349}{100} + \frac{594}{10} + 27$ Let's use a common denominator of 1000: $f(-\frac{9}{10}) = -\frac{7290}{1000} + \frac{23490}{1000} + \frac{59400}{1000} + \frac{27000}{1000}$ $f(-\frac{9}{10}) = \frac{-7290 + 23490 + 59400 + 27000}{1000}$ $f(-\frac{9}{10}) = \frac{102600}{1000} = 102.6$

Since $f(-\frac{9}{10}) \neq 0$, option B is not an actual root. Phew, okay, moving on!

Testing Option C: $\frac{3}{5}$

Let's plug in $x = 3/5$: $f(\frac{3}{5}) = 10(\frac{3}{5})^3 + 29(\frac{3}{5})^2 - 66(\frac{3}{5}) + 27$ $f(\frac{3}{5}) = 10(\frac{27}{125}) + 29(\frac{9}{25}) - 66(\frac{3}{5}) + 27$ $f(\frac{3}{5}) = \frac{270}{125} + \frac{261}{25} - \frac{198}{5} + 27$ Using a common denominator of 125: $f(\frac{3}{5}) = \frac{270}{125} + \frac{261 \times 5}{125} - \frac{198 \times 25}{125} + \frac{27 \times 125}{125}$ $f(\frac{3}{5}) = \frac{270}{125} + \frac{1305}{125} - \frac{4950}{125} + \frac{3375}{125}$ $f(\frac{3}{5}) = \frac{270 + 1305 - 4950 + 3375}{125}$ $f(\frac{3}{5}) = \frac{4950 - 4950}{125}$ $f(\frac{3}{5}) = \frac{0}{125} = 0$

Awesome! Option C, $\frac{3}{5}$, is also an actual root. We're on a roll, guys!

Testing Option D: 1

Let's test $x = 1$: $f(1) = 10(1)^3 + 29(1)^2 - 66(1) + 27$ $f(1) = 10(1) + 29(1) - 66 + 27$ $f(1) = 10 + 29 - 66 + 27$ $f(1) = 39 - 66 + 27$ $f(1) = -27 + 27$ $f(1) = 0$

Bingo! Option D, 1, is another actual root. This is fantastic!

Testing Option E: 3

Finally, let's check $x = 3$: $f(3) = 10(3)^3 + 29(3)^2 - 66(3) + 27$ $f(3) = 10(27) + 29(9) - 198 + 27$ $f(3) = 270 + 261 - 198 + 27$ $f(3) = 531 - 198 + 27$ $f(3) = 333 + 27$ $f(3) = 360$

Since $f(3) \neq 0$, option E is not an actual root. Too bad!

Conclusion: Identifying the Actual Roots

After diligently testing each potential root using the Rational Root Theorem and direct substitution, we found that three of the options made our polynomial $f(x)=10 x^3+29 x^2-66 x+27$ equal to zero. These are the actual roots of the polynomial from the given list. Remember, the Rational Root Theorem is your best friend for finding potential rational roots, but you always have to verify them by plugging them back into the function. It's a two-step process that ensures accuracy. The power of mathematics lies in these systematic approaches that break down complex problems into manageable steps. By understanding the theorem and patiently applying it, we've successfully identified the roots. It’s a great feeling to conquer these problems, right?

Therefore, the actual roots from the given options are:

• A. $-\frac{9}{2}$
• C. $\frac{3}{5}$
• D. 1

These three values are the ones that satisfy the equation $f(x)=0$. Keep practicing, and you'll become a polynomial root-finding pro in no time! Math is all about understanding the tools you have and knowing how to use them effectively. The Rational Root Theorem is definitely one of those essential tools in your algebra arsenal. It’s a foundational concept that opens doors to solving more complex equations and understanding the behavior of functions. So next time you see a polynomial, don't shy away – embrace the challenge and let the Rational Root Theorem guide you to the solution! It’s not just about finding answers; it’s about the journey of discovery and the logical steps that lead you there. Happy solving, everyone!