Find The Actual Roots Of Polynomial F(x) Using Rational Root Theorem

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the fascinating world of polynomials and how to find their roots. If you're a math whiz or just trying to wrap your head around calculus concepts, you're in the right place. We'll be tackling a specific problem that involves the Rational Root Theorem, a super handy tool for locating potential rational roots of a polynomial. So, grab your calculators, maybe a snack, and let's get started on uncovering which of the given options is an actual root of the polynomial $f(x)=6 x^4+5 x^3-33 x^2-12 x+20$. We've got a list of potential candidates: -rac{5}{2}, $-2$, $1$, and rac{10}{3}. The goal is to determine which one actually makes the polynomial equal to zero when substituted in. This process is all about testing these potential roots, and the Rational Root Theorem helps us narrow down the possibilities, making our lives a whole lot easier. Let's break down how we can use this theorem and then test each option systematically to find the true root. Get ready to flex those math muscles!

Understanding the Rational Root Theorem

The Rational Root Theorem is a cornerstone in algebra when dealing with polynomials. It provides a systematic way to find all possible rational roots of a polynomial equation with integer coefficients. For a polynomial $f(x) = a_n x^n + a_{n-1} x^{n-1} + ext{...} + a_1 x + a_0$, where all coefficients ($a_i$) are integers, the theorem states that any rational root, expressible as rac{p}{q} (where $p$ and $q$ are integers with no common factors other than 1, and $q eq 0$), must satisfy two conditions: $p$ must be a divisor of the constant term ($a_0$), and $q$ must be a divisor of the leading coefficient ($a_n$). In simpler terms, you take all the factors of the last number (the constant term) and divide them by all the factors of the first number (the coefficient of the highest power of $x$). This gives you a list of potential rational roots. It's crucial to remember that this theorem only gives us potential roots; we still need to test them to see which ones are actual roots.

In our specific problem, the polynomial is $f(x)=6 x^4+5 x^3-33 x^2-12 x+20$. Here, the constant term ($a_0$) is $20$, and the leading coefficient ($a_n$, which is $a_4$ in this case) is $6$. According to the Rational Root Theorem, any rational root rac{p}{q} must have $p$ as a factor of $20$ and $q$ as a factor of $6$.

Let's list the factors of $20$: $\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20$. And the factors of $6$: $\pm 1, \pm 2, \pm 3, \pm 6$.

Now, we form all possible fractions rac{p}{q} by taking each factor of $20$ and dividing it by each factor of $6$. This would give us a comprehensive list of potential rational roots. For example, rac{\pm 1}{\pm 1} = \pm 1, rac{2}{\pm 3} = \pm rac{2}{3}, rac{5}{\pm 6} = \pm rac{5}{6}, and so on. The given options in the question are -rac{5}{2}, $-2$, $1$, and rac{10}{3}. We can verify that these are indeed among the potential rational roots derived from the theorem. For instance, for -rac{5}{2}, $p=-5$ (a factor of $20$) and $q=2$ (a factor of $6$). For $-2$, we can write it as -rac{2}{1}, where $p=-2$ (factor of $20$) and $q=1$ (factor of $6$). For $1$, we can write it as rac{1}{1}, where $p=1$ (factor of $20$) and $q=1$ (factor of $6$). And for rac{10}{3}, $p=10$ (factor of $20$) and $q=3$ (factor of $6$). So, the theorem has correctly identified these as plausible candidates.

Testing the Potential Roots

Now that we've got our potential rational roots, the next crucial step is to test each one to see if it makes the polynomial $f(x)$ equal to zero. This is the process of finding the actual roots. An actual root, or a zero of the polynomial, is a value of $x$ for which $f(x) = 0$. We will substitute each of the given options into the polynomial expression and evaluate. If the result is $0$, then that option is an actual root. Let's roll up our sleeves and test each one systematically.

Testing Option A: x = -rac{5}{2}

We start with option A, x = -rac{5}{2}. We need to substitute this value into $f(x)=6 x^4+5 x^3-33 x^2-12 x+20$ and see if we get $0$.

$f \left(-\frac{5}{2}\right) = 6\left(-\frac{5}{2}\right)^4 + 5\left(-\frac{5}{2}\right)^3 - 33\left(-\frac{5}{2}\right)^2 - 12\left(-\frac{5}{2}\right) + 20$

Let's calculate each term:

$\left(-\frac{5}{2}\right)^4 = \frac{(-5)^4}{2^4} = \frac{625}{16}$

$6\left(\frac{625}{16}\right) = \frac{3750}{16} = \frac{1875}{8}$

$\left(-\frac{5}{2}\right)^3 = \frac{(-5)^3}{2^3} = \frac{-125}{8}$

$5\left(-\frac{125}{8}\right) = -\frac{625}{8}$

$\left(-\frac{5}{2}\right)^2 = \frac{(-5)^2}{2^2} = \frac{25}{4}$

$-33\left(\frac{25}{4}\right) = -\frac{825}{4}$

$-12\left(-\frac{5}{2}\right) = \frac{60}{2} = 30$

Now, let's combine everything:

$f\left(-\frac{5}{2}\right) = \frac{1875}{8} - \frac{625}{8} - \frac{825}{4} + 30 + 20$

To add these fractions, we need a common denominator, which is $8$.

$f\left(-\frac{5}{2}\right) = \frac{1875}{8} - \frac{625}{8} - \frac{825 \times 2}{4 \times 2} + 50$

$f\left(-\frac{5}{2}\right) = \frac{1875}{8} - \frac{625}{8} - \frac{1650}{8} + \frac{50 \times 8}{8}$

$f\left(-\frac{5}{2}\right) = \frac{1875 - 625 - 1650 + 400}{8}$

$f\left(-\frac{5}{2}\right) = \frac{1250 - 1650 + 400}{8}$

$f\left(-\frac{5}{2}\right) = \frac{-400 + 400}{8} = \frac{0}{8} = 0$

Wow! It looks like -rac{5}{2} is indeed an actual root because $f \left(-\frac{5}{2}\right) = 0$. This means option A is our answer. However, to be thorough and to solidify our understanding, let's test the other options as well. It's always good practice to double-check, especially in math!

Testing Option B: $x = -2$

Let's test $x = -2$. We substitute $-2$ into $f(x)=6 x^4+5 x^3-33 x^2-12 x+20$.

$f(-2) = 6(-2)^4 + 5(-2)^3 - 33(-2)^2 - 12(-2) + 20$

Calculate each term:

$(-2)^4 = 16$

$6(16) = 96$

$(-2)^3 = -8$

$5(-8) = -40$

$(-2)^2 = 4$

$-33(4) = -132$

$-12(-2) = 24$

Now, sum them up:

$f(-2) = 96 - 40 - 132 + 24 + 20$

$f(-2) = 56 - 132 + 24 + 20$

$f(-2) = -76 + 24 + 20$

$f(-2) = -52 + 20$

$f(-2) = -32$

Since $f(-2) = -32 \neq 0$, $x = -2$ is not an actual root. So, option B is incorrect.

Testing Option C: $x = 1$

Next, we test $x = 1$. This is usually the easiest one to check.

$f(1) = 6(1)^4 + 5(1)^3 - 33(1)^2 - 12(1) + 20$

$f(1) = 6(1) + 5(1) - 33(1) - 12(1) + 20$

$f(1) = 6 + 5 - 33 - 12 + 20$

$f(1) = 11 - 33 - 12 + 20$

$f(1) = -22 - 12 + 20$

$f(1) = -34 + 20$

$f(1) = -14$

Since $f(1) = -14 \neq 0$, $x = 1$ is not an actual root. So, option C is incorrect.

Testing Option D: x = rac{10}{3}

Finally, let's test option D, x = rac{10}{3}. This one involves larger numbers, so let's be careful with our calculations.

$f\left(\frac{10}{3}\right) = 6\left(\frac{10}{3}\right)^4 + 5\left(\frac{10}{3}\right)^3 - 33\left(\frac{10}{3}\right)^2 - 12\left(\frac{10}{3}\right) + 20$

Calculate each term:

$\left(\frac{10}{3}\right)^4 = \frac{10^4}{3^4} = \frac{10000}{81}$

$6\left(\frac{10000}{81}\right) = \frac{60000}{81} = \frac{20000}{27}$

$\left(\frac{10}{3}\right)^3 = \frac{10^3}{3^3} = \frac{1000}{27}$

$5\left(\frac{1000}{27}\right) = \frac{5000}{27}$

$\left(\frac{10}{3}\right)^2 = \frac{10^2}{3^2} = \frac{100}{9}$

$-33\left(\frac{100}{9}\right) = -\frac{3300}{9} = -\frac{1100}{3}$

$-12\left(\frac{10}{3}\right) = -\frac{120}{3} = -40$

Now, let's combine everything:

$f\left(\frac{10}{3}\right) = \frac{20000}{27} + \frac{5000}{27} - \frac{1100}{3} - 40 + 20$

$f\left(\frac{10}{3}\right) = \frac{25000}{27} - \frac{1100}{3} - 20$

To get a common denominator (which is $27$):

$f\left(\frac{10}{3}\right) = \frac{25000}{27} - \frac{1100 \times 9}{3 \times 9} - \frac{20 \times 27}{27}$

$f\left(\frac{10}{3}\right) = \frac{25000}{27} - \frac{9900}{27} - \frac{540}{27}$

$f\left(\frac{10}{3}\right) = \frac{25000 - 9900 - 540}{27}$

$f\left(\frac{10}{3}\right) = \frac{15100 - 540}{27}$

$f\left(\frac{10}{3}\right) = \frac{14560}{27}$

Since $\frac{14560}{27} \neq 0$, $x = \frac{10}{3}$ is not an actual root. So, option D is incorrect.

Conclusion: The Actual Root

After meticulously testing all the given options using the Rational Root Theorem and direct substitution, we found that only one value resulted in $f(x) = 0$. That value is x = -rac{5}{2}. This confirms that option A is the correct answer. It's pretty cool how the Rational Root Theorem gives us a manageable list of candidates, and then a bit of careful calculation reveals the true solutions. Remember, practice makes perfect, so keep working through these polynomial problems. You've got this!

The final answer is $\boxed{-\frac{5}{2}}$.