Finding Roots Of Polynomial Functions: A Comprehensive Guide

Hey everyone! Today, we're diving into a cool concept in algebra: finding the roots of polynomial functions. We'll break down the problem step-by-step and then look at the answer choices to see what makes the most sense. This is a topic that can seem a little tricky at first, but trust me, with a little practice, you'll get the hang of it. So, let's get started!

Understanding the Basics: Polynomial Roots and Rational Coefficients

Okay, guys, first things first: What does it all mean? We're talking about polynomial functions, which are equations like f(x) = x^2 + 2x + 1. The 'roots' of a polynomial are the values of 'x' that make the function equal to zero. Think of it like finding the x-intercepts on a graph; those are the points where the graph crosses the x-axis, and they correspond to the roots. Now, in this particular question, we are given that the polynomial has rational coefficients. This is a super important detail, so make sure you don't miss it. Rational coefficients means the numbers multiplying the 'x' terms are all rational numbers (numbers that can be expressed as a fraction, like 1/2 or 3/4). Why is this relevant? Because it dictates the behavior of the roots, especially when we start dealing with irrational numbers or complex numbers.

Now, let's talk about the specific roots we know. We are told the function has roots at 0, 4, and 3 + √11. The presence of 3 + √11 is where things get interesting. √11 is an irrational number (a number that can't be expressed as a simple fraction). Because our polynomial has rational coefficients, if a root contains a radical (like √11), then its conjugate must also be a root. This is a fundamental rule that helps us solve these types of problems. The conjugate of an expression a + b√c is a - b√c. This rule is super helpful because it means we don't have to do a ton of extra calculations. Also, this concept applies to complex numbers too! If a complex number, like a + bi, is a root of a polynomial with rational coefficients, then its conjugate, a - bi, must also be a root. It's like having a special pair of twins within the world of roots.

Conjugate Root Theorem: The Key to Solving This Problem

Alright, let's get down to the meat of this problem. The Conjugate Root Theorem states that if a polynomial with rational coefficients has an irrational root of the form a + b√c, then it must also have a root of the form a - b√c. This is because when we solve for roots using methods like the quadratic formula, irrational roots always come in pairs. This concept simplifies our search for other roots. In this scenario, we know the polynomial has rational coefficients and one of the roots is 3 + √11. Using the conjugate root theorem, we know immediately that the conjugate of 3 + √11 must also be a root. So, what's the conjugate of 3 + √11? It's simply 3 - √11. That's the key to finding the answer! The conjugate root theorem basically tells us the other root is the same as the given root but with the sign changed between the rational and irrational parts. It is a fundamental rule for polynomials with rational coefficients. It saves us time and effort and makes sure we're always on the right track. This principle extends to complex roots, meaning that if a polynomial has complex roots, the complex conjugates will always be roots too.

Analyzing the Answer Choices:

Now, let's go through the answer choices to see which one fits the bill and reinforces our learnings from the previous two sections. In this section, we'll see if we can identify the correct choice right away. By doing this, we also learn how to identify other possible answers that may be very common misconceptions that students may have when encountering this type of problem.

A. 3 + √11: This is the original root given in the problem, not its conjugate. Therefore, this isn't the correct choice because, according to the conjugate root theorem, we know we should be looking for the conjugate, which has a sign change. In short, the presence of 3 + √11 tells us the conjugate, 3 - √11, should also be there, not 3 + √11. So, we're not quite there yet!

B. -3 + i√11: This choice involves a complex number (the 'i' represents the imaginary unit, the square root of -1), but our given root is real (3 + √11 is a real number, even though √11 is irrational). Since our roots are not in a complex format, we won't be using this choice. This means we are only focusing on irrational and rational numbers.

C. 3 - √11: This is it! This is the conjugate of 3 + √11. As we discussed earlier, if 3 + √11 is a root, the conjugate root theorem tells us that 3 - √11 must also be a root. It perfectly matches our explanation.

D. -3 - √11: This answer might appear similar, but it's not the conjugate of 3 + √11. The conjugate root theorem doesn't just change the sign of the whole expression; it changes the sign in front of the irrational part while keeping the sign of the rational part the same. This is where most students get tripped up.

Conclusion: Finding the Right Root

So, after a good look at all the answer options, the correct answer is indeed C. 3 - √11. Because the polynomial has rational coefficients, and we are given that 3 + √11 is a root, we know that its conjugate, 3 - √11, must also be a root, thanks to the Conjugate Root Theorem. This concept is a cornerstone in understanding the behavior of polynomial functions. It shows us how roots come in pairs, which is especially important when dealing with irrational and complex numbers. Keep in mind that understanding and mastering the Conjugate Root Theorem is essential to success in algebra and beyond. I hope this explanation has been helpful. Keep up the great work! If you have any questions, feel free to ask! Understanding the core concepts and applying the theorems is important to getting the right answer!