Finding Roots Of Polynomial Functions A Step By Step Guide

Hey guys! Let's dive into the fascinating world of polynomial functions, specifically focusing on how to find their roots. Today, we're tackling the polynomial function $F(x) = 2x^3 - 5x^2 + 2x + 1$. Our mission? To identify which of the given options—$\frac{5+\sqrt{10}}{6}$, 1, $\frac{3-\sqrt{17}}{4}$, and $\frac{3+\sqrt{17}}{4}$—are indeed roots of this function. So, buckle up, and let’s get started!

Understanding Polynomial Roots

Before we jump into the nitty-gritty, let's make sure we're all on the same page about what a root actually is. In simple terms, a root of a polynomial function is a value of x that makes the function equal to zero. Mathematically, if r is a root of F(x), then F(r) = 0. This is a crucial concept because finding roots helps us understand where the graph of the polynomial intersects the x-axis. These intersections are also known as the x-intercepts, which are key features of the polynomial’s behavior and shape. When we talk about solving for the roots, we're essentially finding the values of x that satisfy the equation $2x^3 - 5x^2 + 2x + 1 = 0$. Now, this isn't always a straightforward process, especially with cubic polynomials like the one we have here. There isn't a simple, one-size-fits-all formula like the quadratic formula for finding roots of quadratic equations. Instead, we often rely on a mix of techniques, including factoring (if possible), the Rational Root Theorem, synthetic division, and sometimes, good old-fashioned trial and error. Factoring is a brilliant way to simplify the problem if we can break down the polynomial into smaller, more manageable factors. For example, if we could rewrite our cubic polynomial as a product of linear factors like $(x - r_1)(x - r_2)(x - r_3)$, then the roots would simply be $r_1$, $r_2$, and $r_3$. However, not all polynomials can be factored easily, especially over the rational numbers. That's where the Rational Root Theorem comes in handy. It provides a list of potential rational roots, which we can then test using synthetic division or direct substitution. The Rational Root Theorem tells us that if a polynomial has rational roots (roots that can be expressed as a fraction), they will be of the form $\pm\frac{p}{q}$, where p is a factor of the constant term and q is a factor of the leading coefficient. Synthetic division is a streamlined method for dividing a polynomial by a linear factor of the form $(x - c)$. If the remainder after synthetic division is zero, then c is a root of the polynomial. This method is not only efficient for testing potential roots but also helps reduce the degree of the polynomial, making it easier to find the remaining roots. For instance, if we find one root of our cubic polynomial using synthetic division, we can reduce it to a quadratic, which we can then solve using the quadratic formula or factoring.

Testing the Potential Roots

Alright, let's roll up our sleeves and get to the heart of the problem. We need to figure out which of the given options are actual roots of the polynomial function $F(x) = 2x^3 - 5x^2 + 2x + 1$. We have four candidates: $\frac5+\sqrt{10}}{6}$, 1, $\frac{3-\sqrt{17}}{4}$, and $\frac{3+\sqrt{17}}{4}$. The most direct way to check if a number is a root is to substitute it into the function and see if we get zero. It might sound tedious, but it's a reliable method. Let's start with option B, which is 1. This one looks promising because it's a simple integer, and these are often the easiest to work with. We'll substitute x = 1 into our function $F(1) = 2(1)^3 - 5(1)^2 + 2(1) + 1 = 2 - 5 + 2 + 1 = 0$. Bingo! It turns out that $F(1) = 0$, so 1 is indeed a root of our polynomial. Now, let's tackle the other options. Options A, C, and D involve radicals, which means substituting them directly might get a bit messy. However, we can still use the same principle: if the result is zero, then the number is a root. Let’s try option D, $\frac{3+\sqrt{17}4}$. Substituting this into the function gives us $F\left(\frac{3+\sqrt{17}4}\right) = 2\left(\frac{3+\sqrt{17}}{4}\right)^3 - 5\left(\frac{3+\sqrt{17}}{4}\right)^2 + 2\left(\frac{3+\sqrt{17}}{4}\right) + 1$. This looks like a beast to calculate by hand, and honestly, it is! But if we carefully evaluate each term, we'll find that it simplifies to zero. This confirms that $\frac{3+\sqrt{17}}{4}$ is also a root. Now, what about option C, $\frac{3-\sqrt{17}}{4}$? Notice anything familiar? This looks like the conjugate of option D. In many cases, when dealing with polynomials with rational coefficients, irrational roots often come in conjugate pairs. This means that if $a + b\sqrt{c}$ is a root, then $a - b\sqrt{c}$ is also likely to be a root. Let's substitute $\frac{3-\sqrt{17}}{4}$ into the function $F\left(\frac{3-\sqrt{17}4}\right) = 2\left(\frac{3-\sqrt{17}}{4}\right)^3 - 5\left(\frac{3-\sqrt{17}}{4}\right)^2 + 2\left(\frac{3-\sqrt{17}}{4}\right) + 1$. Just like with option D, this calculation is lengthy, but it simplifies to zero as well. So, $\frac{3-\sqrt{17}}{4}$ is another root. Finally, let's consider option A, $\frac{5+\sqrt{10}}{6}$. Substituting this into the function $F\left(\frac{5+\sqrt{10}{6}\right) = 2\left(\frac{5+\sqrt{10}}{6}\right)^3 - 5\left(\frac{5+\sqrt{10}}{6}\right)^2 + 2\left(\frac{5+\sqrt{10}}{6}\right) + 1$. Evaluating this expression is quite involved, and unlike the previous cases, it does not simplify to zero. Therefore, $\frac{5+\sqrt{10}}{6}$ is not a root of the polynomial.

Utilizing Synthetic Division and the Rational Root Theorem

Now that we've tested each option by direct substitution, let's take a step back and explore how we could have approached this problem using other techniques like synthetic division and the Rational Root Theorem. These tools can be incredibly helpful, especially when you're facing a polynomial with higher degrees or when the roots aren't immediately obvious. Synthetic division, as we mentioned earlier, is a streamlined way to divide a polynomial by a linear factor. It’s particularly useful for testing potential rational roots. Remember, if the remainder is zero after synthetic division, then the tested value is a root. Let's put this into action with our polynomial $F(x) = 2x^3 - 5x^2 + 2x + 1$. We already know that 1 is a root, so let's use synthetic division to divide the polynomial by $(x - 1)$: 1 | 2 -5 2 1 | 2 -3 -1 ---------------- 2 -3 -1 0 The numbers in the bottom row (2, -3, -1) represent the coefficients of the quotient polynomial, which is $2x^2 - 3x - 1$. The last number, 0, is the remainder, confirming that 1 is indeed a root. Now, we've reduced our cubic polynomial to a quadratic, which is much easier to handle. We can find the remaining roots by solving the quadratic equation $2x^2 - 3x - 1 = 0$. We can use the quadratic formula for this: $x = \frac-b \pm \sqrt{b^2 - 4ac}}{2a}$. Plugging in the coefficients a = 2, b = -3, and c = -1, we get $x = \frac{3 \pm \sqrt{(-3)^2 - 4(2)(-1)}{2(2)} = \frac{3 \pm \sqrt{9 + 8}}{4} = \frac{3 \pm \sqrt{17}}{4}$. These are exactly the roots we identified earlier as options C and D! This demonstrates the power of synthetic division in simplifying the problem and making it easier to find the roots. But what if we hadn't known that 1 was a root to begin with? This is where the Rational Root Theorem shines. This theorem gives us a list of potential rational roots to test. For our polynomial $F(x) = 2x^3 - 5x^2 + 2x + 1$, the constant term is 1, and the leading coefficient is 2. The factors of the constant term (1) are $\pm 1$, and the factors of the leading coefficient (2) are $\pm 1$ and $\pm 2$. Therefore, the potential rational roots are $\pm \frac{1}{1}$ and $\pm \frac{1}{2}$, which simplifies to $\pm 1$ and $\pm \frac{1}{2}$. We can then use synthetic division or direct substitution to test these values. We already saw that 1 is a root. If we tested -1, $\frac{1}{2}$, and -$\frac{1}{2}$, we'd find that none of them are roots, which means the other roots must be irrational or complex. This would lead us to suspect the radical options (C and D) and test them accordingly.

Conclusion: Identifying the Roots

Alright guys, we've reached the finish line! We've thoroughly explored the polynomial function $F(x) = 2x^3 - 5x^2 + 2x + 1$ and successfully identified its roots. Through direct substitution, we confirmed that 1, $\frac{3-\sqrt{17}}{4}$, and $\frac{3+\sqrt{17}}{4}$ are indeed roots of the function. We also learned that $\frac{5+\sqrt{10}}{6}$ is not a root. Additionally, we delved into the power of synthetic division and the Rational Root Theorem, which are invaluable tools for tackling polynomial root-finding problems. Synthetic division allowed us to reduce the cubic polynomial to a quadratic after finding one root, making it easier to find the remaining roots using the quadratic formula. The Rational Root Theorem provided us with a systematic way to identify potential rational roots, which we could then test using synthetic division or direct substitution. So, to wrap it all up, the roots of the polynomial function $F(x) = 2x^3 - 5x^2 + 2x + 1$ from the given options are:

• B. 1
• C. $\frac{3-\sqrt{17}}{4}$
• D. $\frac{3+\sqrt{17}}{4}$

Remember, finding roots is a fundamental concept in polynomial functions and has wide applications in mathematics and beyond. Whether you're sketching graphs, solving equations, or modeling real-world phenomena, understanding how to find roots is a key skill. Keep practicing, and you'll become a root-finding pro in no time!