Find Polynomial With 11 Roots: Algebra Theorem Guide

Aug 6, 2025 by ADMIN 53 views

Understanding Polynomial Roots: Finding the 11-Root Polynomial

Hey guys! Let's dive into the fascinating world of polynomials and their roots. Today, we're tackling a question rooted in the Fundamental Theorem of Algebra: Which polynomial function has exactly 11 roots? This theorem is a cornerstone of algebra, so understanding it is crucial for anyone delving into higher mathematics. We'll break down the theorem, explore the options, and pinpoint the correct answer. Buckle up, it's gonna be an educational ride!

The Fundamental Theorem of Algebra: A Quick Recap

Before we jump into the problem, let's refresh our understanding of the Fundamental Theorem of Algebra. In simple terms, this theorem states that a polynomial equation of degree n has exactly n complex roots, counting multiplicities. What does this mean?

Degree of a Polynomial: The degree of a polynomial is the highest power of the variable (usually x) in the polynomial. For example, in the polynomial $x^3 + 2x^2 - x + 5$, the degree is 3.
Roots of a Polynomial: The roots of a polynomial are the values of the variable that make the polynomial equal to zero. These are also sometimes called zeros or solutions of the polynomial equation.
Complex Roots: Complex roots include both real numbers and imaginary numbers (numbers involving the imaginary unit i, where $i^2 = -1$). So, every real number is also a complex number (with an imaginary part of zero).
Multiplicity of a Root: The multiplicity of a root is the number of times that root appears as a solution to the polynomial equation. For instance, in the polynomial $(x-2)^3$, the root 2 has a multiplicity of 3 because the factor $(x-2)$ appears three times.

So, the Fundamental Theorem of Algebra essentially guarantees that a polynomial of degree n will have n roots, some of which may be repeated (multiplicity) and some of which may be complex. This is a super powerful concept, and it's the key to solving our problem. To truly grasp this, let's delve a bit deeper with an example. Consider the quadratic equation $x^2 - 4x + 4 = 0$. This polynomial has a degree of 2. We can factor this equation as $(x-2)(x-2) = 0$, which means it has two roots, both equal to 2. Thus, the root 2 has a multiplicity of 2. This equation perfectly illustrates the theorem: a polynomial of degree 2 has 2 roots, counting multiplicity. The beauty of the Fundamental Theorem of Algebra is its guarantee. We know, without even solving the polynomial, how many roots to expect. This is invaluable in many areas of mathematics and engineering. Now, let’s apply this understanding to the given options and determine which polynomial has exactly 11 roots.

Analyzing the Options: Finding the Polynomial with 11 Roots

Now that we've got a solid grasp of the Fundamental Theorem of Algebra, let's apply it to our problem. We need to identify the polynomial function that has exactly 11 roots. Remember, we're looking for the polynomial whose degree (when fully expanded) equals 11, considering the multiplicities of the roots.

Let's examine each option:

A. $f(x)=(x-1)(x+1)^{11}$

Here, we have two factors: $(x-1)$ and $(x+1)^{11}$. The factor $(x-1)$ has a degree of 1, and the factor $(x+1)^{11}$ has a degree of 11. When we multiply these factors, we add their degrees. So, the degree of the entire polynomial is $1 + 11 = 12$. Therefore, this polynomial has 12 roots, not 11. We can also see that the root $x = 1$ has a multiplicity of 1, and the root $x = -1$ has a multiplicity of 11. The sum of the multiplicities, 1 + 11, gives us the total number of roots, which is 12. Thus, option A is not the answer.

B. $f(x)=(x+2)^3[0.2cm] \left(x^2-7 x+3\right)^4$

This polynomial has two factors: $(x+2)^3$ and $(x^2 - 7x + 3)^4$. The factor $(x+2)^3$ has a degree of 3. The factor $(x^2 - 7x + 3)$ is a quadratic (degree 2), and it's raised to the power of 4, so its degree is $2 * 4 = 8$. The total degree of the polynomial is $3 + 8 = 11$. Bingo! This polynomial has a degree of 11 and, according to the Fundamental Theorem of Algebra, has exactly 11 roots. However, let's not jump to conclusions just yet. We need to examine the other options to make sure this is indeed the correct answer. The root $x = -2$ has a multiplicity of 3. The quadratic factor $(x^2 - 7x + 3)^4$ will have 4 pairs of roots (since a quadratic has 2 roots), each with a multiplicity of 4, contributing 8 roots in total. Adding these up, we have 3 + 8 = 11 roots, confirming our suspicion.

C. $f(x)=\left(x^5+7 x+14\right)^6$

Here, we have a polynomial of degree 5, $(x^5 + 7x + 14)$, raised to the power of 6. The degree of the entire polynomial is $5 * 6 = 30$. This polynomial has 30 roots, far more than our target of 11. So, this option is incorrect. The quintic polynomial $(x^5 + 7x + 14)$ has 5 roots, and since it's raised to the power of 6, each of those 5 roots has a multiplicity of 6, giving us a total of 30 roots.

D. $f(x)=11$

This is a constant function. It has a degree of 0 (since there's no x term). Therefore, it has 0 roots. This option is definitely not the answer.

After carefully analyzing each option, it's clear that Option B is the correct answer. The polynomial $f(x)=(x+2)^3(x2-7x+3)^4$ has a degree of 11 and, consequently, has exactly 11 roots, considering multiplicities. This exercise highlights the power of the Fundamental Theorem of Algebra in quickly determining the number of roots of a polynomial.

The Verdict: Option B is the Winner!

Alright, guys, we've cracked the code! By applying the Fundamental Theorem of Algebra and meticulously analyzing each option, we've confidently determined that the polynomial function with exactly 11 roots is:

B. $f(x)=(x+2)^3(x2-7x+3)^4$

This question perfectly illustrates how a strong understanding of fundamental theorems can simplify seemingly complex problems. Remember, the degree of the polynomial (accounting for multiplicities) dictates the number of roots. Keep practicing, and these concepts will become second nature. Now, let's move on to even more exciting mathematical adventures!