Finding Roots Of Polynomial Function F(x)=2x^3-5x^2+2x+1

Determining the roots of a polynomial function is a fundamental problem in algebra and calculus. The roots, also known as zeros, of a polynomial $F(x)$ are the values of $x$ for which $F(x) = 0$. These roots provide crucial information about the behavior of the function, such as where it intersects the x-axis, its turning points, and its overall shape. In this detailed exploration, we will delve into the process of finding the roots of the polynomial function $F(x) = 2x^3 - 5x^2 + 2x + 1$. We will examine several techniques, including the Rational Root Theorem, synthetic division, and the quadratic formula, to identify the roots among the given options: A. 1, B. $\frac{3+\sqrt{17}}{4}$, C. $\frac{5-\sqrt{10}}{6}$, and D. $\frac{3-\sqrt{17}}{4}$. This guide will not only provide a step-by-step solution but also offer insights into the underlying mathematical principles and strategies for solving similar problems.

Understanding Polynomial Roots

Before we dive into the specific problem, it's crucial to grasp the concept of polynomial roots and their significance. A root of a polynomial function $F(x)$ is a value $x = r$ such that $F(r) = 0$. Graphically, these roots correspond to the points where the polynomial's graph intersects the x-axis. Finding the roots is essential for various applications, including solving equations, analyzing the behavior of functions, and modeling real-world phenomena. For a polynomial of degree $n$, there are at most $n$ roots, considering both real and complex roots, and counting multiplicities. This is a consequence of the Fundamental Theorem of Algebra, which states that every non-constant single-variable polynomial with complex coefficients has at least one complex root. The roots can be rational, irrational, or complex, and they provide a complete picture of the polynomial's zero crossings.

The roots of a polynomial are intimately connected with its factors. If $r$ is a root of $F(x)$, then $(x - r)$ is a factor of $F(x)$. Conversely, if $(x - r)$ is a factor of $F(x)$, then $r$ is a root. This relationship is formalized in the Factor Theorem, which is a powerful tool for finding and verifying roots. For instance, if we find that $x = 1$ is a root of $F(x)$, then $(x - 1)$ must be a factor of $F(x)$. This fact can be used to simplify the polynomial by dividing $F(x)$ by $(x - 1)$, resulting in a polynomial of lower degree that is easier to analyze. Understanding this relationship between roots and factors is crucial for solving polynomial equations and for understanding the structure of polynomials.

Furthermore, the Rational Root Theorem is a valuable tool for identifying 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. This theorem provides a finite list of possible rational roots that can be tested using synthetic division or direct substitution. While the Rational Root Theorem does not guarantee the existence of rational roots, it significantly narrows down the search and can often lead to the discovery of roots that might otherwise be difficult to find. In the context of our problem, we will use the Rational Root Theorem to identify potential rational roots of $F(x) = 2x^3 - 5x^2 + 2x + 1$ before exploring other methods for finding roots.

Applying the Rational Root Theorem to $F(x)=2x^3-5x^2+2x+1$

The Rational Root Theorem provides a systematic way to identify potential rational roots of a polynomial. For our polynomial $F(x) = 2x^3 - 5x^2 + 2x + 1$, the constant term is 1, and the leading coefficient is 2. According to the theorem, any rational root $\frac{p}{q}$ must have $p$ as a factor of 1 and $q$ as a factor of 2. The factors of 1 are $\pm 1$, and the factors of 2 are $\pm 1$ and $\pm 2$. Therefore, the possible rational roots are $\pm \frac{1}{1}$ and $\pm \frac{1}{2}$, which simplifies to $\pm 1$ and $\pm \frac{1}{2}$.

We now test these potential roots by substituting them into the polynomial $F(x)$.

1. Testing $x = 1$: $F(1) = 2(1)^3 - 5(1)^2 + 2(1) + 1 = 2 - 5 + 2 + 1 = 0$. Thus, $x = 1$ is a root of $F(x)$.
2. Testing $x = -1$: $F(-1) = 2(-1)^3 - 5(-1)^2 + 2(-1) + 1 = -2 - 5 - 2 + 1 = -8$. Thus, $x = -1$ is not a root.
3. Testing $x = \frac{1}{2}$: $F(\frac{1}{2}) = 2(\frac{1}{2})^3 - 5(\frac{1}{2})^2 + 2(\frac{1}{2}) + 1 = 2(\frac{1}{8}) - 5(\frac{1}{4}) + 1 + 1 = \frac{1}{4} - \frac{5}{4} + 2 = -1 + 2 = 1$. Thus, $x = \frac{1}{2}$ is not a root.
4. Testing $x = -\frac{1}{2}$: $F(-\frac{1}{2}) = 2(-\frac{1}{2})^3 - 5(-\frac{1}{2})^2 + 2(-\frac{1}{2}) + 1 = 2(-\frac{1}{8}) - 5(\frac{1}{4}) - 1 + 1 = -\frac{1}{4} - \frac{5}{4} = -\frac{6}{4} = -\frac{3}{2}$. Thus, $x = -\frac{1}{2}$ is not a root.

From this, we find that $x = 1$ is the only rational root identified by the Rational Root Theorem. This means option A, $x=1$, is one of the roots of the polynomial function. Now that we have confirmed one root, we can use synthetic division to reduce the cubic polynomial to a quadratic, making it easier to find the remaining roots.

Using Synthetic Division to Reduce the Polynomial

Since we have determined that $x = 1$ is a root of $F(x) = 2x^3 - 5x^2 + 2x + 1$, we can use synthetic division to divide $F(x)$ by $(x - 1)$. Synthetic division is a streamlined method for dividing a polynomial by a linear factor of the form $(x - r)$, where $r$ is a constant. It simplifies the division process and allows us to efficiently find the quotient and remainder. In this case, we will divide $F(x)$ by $(x - 1)$ to reduce the cubic polynomial to a quadratic polynomial.

To perform synthetic division, we set up a table with the coefficients of the polynomial (2, -5, 2, 1) and the root we are dividing by (1).

1. Bring down the first coefficient (2) to the bottom row.

2. Multiply the root (1) by the number in the bottom row (2) and write the result (2) under the next coefficient (-5).

3. Add the numbers in the second column (-5 and 2) and write the sum (-3) in the bottom row.

4. Multiply the root (1) by the number in the bottom row (-3) and write the result (-3) under the next coefficient (2).

5. Add the numbers in the third column (2 and -3) and write the sum (-1) in the bottom row.

6. Multiply the root (1) by the number in the bottom row (-1) and write the result (-1) under the last coefficient (1).

7. Add the numbers in the last column (1 and -1) and write the sum (0) in the bottom row.

The bottom row (2, -3, -1, 0) represents the coefficients of the quotient polynomial and the remainder. Since the remainder is 0, this confirms that $(x - 1)$ is indeed a factor of $F(x)$. The quotient polynomial is $2x^2 - 3x - 1$.

Now, we need to find the roots of the quadratic polynomial $2x^2 - 3x - 1 = 0$. This can be done using the quadratic formula.

Solving the Quadratic Equation Using the Quadratic Formula

The quadratic formula is a universal method for finding the roots of any quadratic equation of the form $ax^2 + bx + c = 0$. The formula is given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For our quadratic equation $2x^2 - 3x - 1 = 0$, we have $a = 2$, $b = -3$, and $c = -1$. Plugging these values into the quadratic formula, we get:

$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-1)}}{2(2)}$

$x = \frac{3 \pm \sqrt{9 + 8}}{4}$

$x = \frac{3 \pm \sqrt{17}}{4}$

Thus, the two roots of the quadratic equation are $x = \frac{3 + \sqrt{17}}{4}$ and $x = \frac{3 - \sqrt{17}}{4}$.

These roots correspond to options B and D. Now we have found all three roots of the original cubic polynomial $F(x)$: 1, $\frac{3 + \sqrt{17}}{4}$, and $\frac{3 - \sqrt{17}}{4}$.

Final Answer

After applying the Rational Root Theorem, synthetic division, and the quadratic formula, we have determined that the roots of the polynomial function $F(x) = 2x^3 - 5x^2 + 2x + 1$ are:

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

Therefore, options A, B, and D are the roots of the polynomial function. This comprehensive approach highlights the interconnectedness of various algebraic techniques in solving polynomial equations, demonstrating the power of combining the Rational Root Theorem, synthetic division, and the quadratic formula.