Rational Root Of Polynomial Function F(x)

Hey guys! Today, we're diving into the fascinating world of polynomial functions, specifically focusing on how to identify potential rational roots. We'll be looking at the polynomial function $f(x) = 5x^5 + \frac{16}{5}x - 3$ and figuring out which value could be a rational root at a particular point, let's call it $P$. Buckle up, because this is going to be an awesome ride!

Understanding Polynomial Functions

Polynomial functions are the bread and butter of algebra, and they come in various forms. The general form of a polynomial function is:

$f(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0$

Where:

• $a_n, a_{n-1}, ..., a_1, a_0$ are coefficients (real numbers).
• $n$ is a non-negative integer (the degree of the polynomial).
• $x$ is the variable.

In our case, we have $f(x) = 5x^5 + \frac{16}{5}x - 3$. This is a fifth-degree polynomial because the highest power of $x$ is 5. Understanding the structure of a polynomial is the first step in finding its roots.

What are Roots?

The roots of a polynomial function are the values of $x$ that make $f(x) = 0$. These are also known as zeros of the function, and they represent the points where the graph of the polynomial intersects the x-axis. Finding these roots is a fundamental problem in algebra, and various techniques exist to solve it.

The Rational Root Theorem

The Rational Root Theorem is a powerful tool that helps us identify potential rational roots of a polynomial. It states that if a polynomial with integer coefficients has a rational root $\frac{p}{q}$ (where $p$ and $q$ are integers with no common factors other than 1), then $p$ must be a factor of the constant term, and $q$ must be a factor of the leading coefficient.

For our polynomial $f(x) = 5x^5 + \frac{16}{5}x - 3$, we first need to make sure we have integer coefficients. To do this, we can multiply the entire function by 5 to get rid of the fraction:

$5f(x) = 25x^5 + 16x - 15$

Now, let's identify the constant term and the leading coefficient:

• Constant term: -15
• Leading coefficient: 25

According to the Rational Root Theorem, any rational root $\frac{p}{q}$ must have $p$ as a factor of -15 and $q$ as a factor of 25.

Factors of -15

The factors of -15 are: $\pm 1, \pm 3, \pm 5, \pm 15$.

Factors of 25

The factors of 25 are: $\pm 1, \pm 5, \pm 25$.

Potential Rational Roots

Now, we can list all possible rational roots by forming fractions $\frac{p}{q}$:

$\pm \frac{1}{1}, \pm \frac{1}{5}, \pm \frac{1}{25}, \pm \frac{3}{1}, \pm \frac{3}{5}, \pm \frac{3}{25}, \pm \frac{5}{1}, \pm \frac{5}{5}, \pm \frac{5}{25}, \pm \frac{15}{1}, \pm \frac{15}{5}, \pm \frac{15}{25}$

Simplifying these fractions, we get:

$\pm 1, \pm \frac{1}{5}, \pm \frac{1}{25}, \pm 3, \pm \frac{3}{5}, \pm \frac{3}{25}, \pm 5, \pm 1, \pm \frac{1}{5}, \pm 15, \pm 3, \pm \frac{3}{5}$

Removing duplicates, our list of potential rational roots is:

$\pm 1, \pm \frac{1}{5}, \pm \frac{1}{25}, \pm 3, \pm \frac{3}{5}, \pm \frac{3}{25}, \pm 5, \pm 15$

Analyzing the Graph at Point P

To determine which of these potential rational roots is likely at point $P$ on the graph, we need to observe the x-coordinate of point $P$. Without the actual graph, we'll make some educated guesses based on common scenarios.

Educated Guesses

1. If Point P is near x = 1:

   If the graph intersects or gets very close to the x-axis at or near $x = 1$, then 1 or -1 could be a rational root. We can test these by plugging them into the polynomial:

   • $f(1) = 5(1)^5 + \frac{16}{5}(1) - 3 = 5 + \frac{16}{5} - 3 = 2 + \frac{16}{5} = \frac{26}{5} \neq 0$
   • $f(-1) = 5(-1)^5 + \frac{16}{5}(-1) - 3 = -5 - \frac{16}{5} - 3 = -8 - \frac{16}{5} = -\frac{56}{5} \neq 0$

   So, neither 1 nor -1 are actual roots, but they remain potential rational roots.

2. If Point P is between 0 and 1:

   If the graph intersects the x-axis between 0 and 1, we should consider fractions like $\frac{1}{5}, \frac{1}{25}, \frac{3}{5}, \frac{3}{25}$. Let's test $\frac{3}{5}$:

   $f(\frac{3}{5}) = 5(\frac{3}{5})^5 + \frac{16}{5}(\frac{3}{5}) - 3$

   This calculation is a bit complex, but it will give us an idea if $\frac{3}{5}$ is close to being a root. If the value is close to zero, it's a good candidate.

3. If Point P is a Larger Value (e.g., x = 3 or x = 5):

   If the graph intersects the x-axis at a larger value, we should consider 3, 5, or 15. These values are less likely for most standard polynomial problems unless specifically designed to have such roots.

Example and Conclusion

Let's pretend that after looking at the graph, point $P$ appears to be very close to $x = \frac{3}{5}$. We would then perform the calculation:

$f(\frac{3}{5}) = 5(\frac{3}{5})^5 + \frac{16}{5}(\frac{3}{5}) - 3$

$f(\frac{3}{5}) = 5(\frac{243}{3125}) + \frac{48}{25} - 3$

$f(\frac{3}{5}) = \frac{243}{625} + \frac{48}{25} - 3$

$f(\frac{3}{5}) = \frac{243}{625} + \frac{1200}{625} - \frac{1875}{625}$

$f(\frac{3}{5}) = \frac{243 + 1200 - 1875}{625} = \frac{-432}{625}$

Since $-\frac{432}{625}$ is not exactly zero, but relatively close, $\frac{3}{5}$ could be a potential rational root that the graph approaches at point $P$.

In conclusion, by using the Rational Root Theorem and observing the graph at point $P$, we can narrow down the list of potential rational roots. Remember, this theorem only gives us potential roots, and we need to verify them by plugging them back into the original polynomial function or using other methods like synthetic division.

Keep exploring those polynomials, guys! You're doing great!