Understanding Polynomial Zeros And Properties

Welcome to our exploration of polynomial properties! Today, we're diving deep into a fascinating problem involving a polynomial named $R(x)$. This polynomial is quite special because it has a degree of 10, meaning the highest power of $x$ in the polynomial is 10. Even more importantly, all of its coefficients are real numbers. This little detail is crucial, as it unlocks some interesting properties we'll be using. We're also given some specific information about its zeros, which are the values of $x$ that make $R(x)$ equal to zero.

We know that $R(x)$ has $4-2i$ as a zero. Now, because the coefficients of $R(x)$ are real numbers, we can use a powerful theorem called the Complex Conjugate Root Theorem. This theorem states that if a polynomial has real coefficients, and if a complex number ($a+bi$) is a zero, then its complex conjugate ($a-bi$) must also be a zero. In our case, the complex zero is $4-2i$. So, its complex conjugate, which is $4+2i$, must also be a zero of $R(x)$. This is our first significant discovery based on the given information. So, when asked to find another zero of $R(x)$, the complex conjugate theorem immediately provides us with $4+2i$ as a definitive answer. This theorem is a cornerstone in understanding the symmetry of zeros in polynomials with real coefficients, ensuring that complex roots always come in pairs.

We are also told that $-2$ is a zero with a multiplicity of 3. What does multiplicity mean? It means that the factor $(x - (-2))$, or simply $(x+2)$, appears three times in the factored form of the polynomial. So, $(x+2)^3$ is a factor of $R(x)$. This tells us that $-2$ is not just a single root, but a repeated root, which impacts the behavior of the polynomial graph at $x=-2$ (it will touch the x-axis and turn around, similar to a parabola's vertex, but potentially with more 'wiggles' depending on the odd multiplicity).

With this information, we already know a few zeros: $4-2i$, $4+2i$, and $-2$ (with multiplicity 3). Let's think about how many zeros we've accounted for so far. We have one complex pair ($4-2i$ and $4+2i$), which counts as two zeros. We also have the zero $-2$ with a multiplicity of 3, which counts as three zeros. So, in total, we have identified $2 + 3 = 5$ zeros. Since the polynomial $R(x)$ has a degree of 10, we know from the Fundamental Theorem of Algebra that it must have exactly 10 zeros, counting multiplicities. This means there are still $10 - 5 = 5$ more zeros to account for.

The problem statement also gives us a hint by asking about the maximum number of real zeros. This is a classic question that relates to the nature of polynomial roots. We know that complex roots (those with a non-zero imaginary part) for polynomials with real coefficients always come in conjugate pairs. This means that any complex zeros will contribute at least two zeros to the total count. If we want to maximize the number of real zeros, we should assume that all the remaining unknown zeros are real. We have 5 remaining zeros to find. If all of them are real numbers, then the maximum number of real zeros would be the 3 we already know (from the multiplicity of -2) plus these 5 potential additional real zeros, totaling $3+5=8$ real zeros. The other 2 zeros would have to be the complex conjugate pair $4-2i$ and $4+2i$. This configuration satisfies the degree of 10 and the conditions given.

Let's summarize what we've deduced. The polynomial $R(x)$ has a degree of 10 and real coefficients. We are given a complex zero $4-2i$. By the Complex Conjugate Root Theorem, its conjugate $4+2i$ must also be a zero. We are also given that $-2$ is a zero with multiplicity 3. So far, we have identified the zeros: $4-2i$, $4+2i$, $-2, -2, -2$. This accounts for 5 of the 10 zeros. The remaining 5 zeros could be any combination of real or complex conjugate pairs. To find the maximum number of real zeros, we assume all remaining 5 zeros are real. Therefore, the maximum number of real zeros would be $3$ (from $-2$) $+ 5$ (new real zeros) $= 8$. The minimum number of real zeros would occur if the remaining 5 zeros were made up of as many complex conjugate pairs as possible. Since we need 5 more zeros, we could have two complex conjugate pairs (4 zeros) and one real zero. This would give us $3$ (from $-2$) $+ 1$ (new real zero) $= 4$ real zeros. So, the number of real zeros can range from 4 to 8.

This problem beautifully illustrates the power of the Complex Conjugate Root Theorem and the Fundamental Theorem of Algebra. These theorems provide essential tools for understanding the structure and behavior of polynomials, especially when dealing with their zeros. By leveraging these mathematical principles, we can uncover hidden information about polynomials, even when not all of their zeros are explicitly given. The interplay between real and complex roots in polynomials with real coefficients is a fundamental concept in algebra, and problems like this offer a practical way to solidify that understanding.

Finding Another Zero

Let's focus on part (a) of the question: Find another zero of $R(x)$. As we discussed earlier, the key piece of information here is that $R(x)$ is a polynomial with real coefficients and that $4-2i$ is one of its zeros. The Complex Conjugate Root Theorem is our direct path to answering this part. This theorem states that if a polynomial has real coefficients, then any complex zeros must occur in conjugate pairs. A complex number is written in the form $a+bi$, where $a$ is the real part and $b$ is the imaginary part. Its complex conjugate is $a-bi$. In our case, the given complex zero is $4-2i$. Here, $a=4$ and $b=-2$. The complex conjugate of $4-2i$ is $4 - (-2)i$, which simplifies to $4+2i$. Therefore, $4+2i$ is guaranteed to be another zero of $R(x)$. This is a direct consequence of the structure imposed by having real coefficients. Without this property, we wouldn't be able to determine any other specific zeros beyond the ones explicitly provided.

So, for part (a), the answer is $4+2i$. This single application of the Complex Conjugate Root Theorem resolves the first part of the problem elegantly. It's a reminder that seemingly abstract theorems have very concrete applications in solving mathematical problems. The existence of this pair of zeros is fundamental to the polynomial's overall structure when its coefficients are restricted to be real numbers.

Maximum Number of Real Zeros

Now, let's tackle part (b): What is the maximum number of real zeros of $R(x)$? We know that $R(x)$ has a degree of 10. The Fundamental Theorem of Algebra tells us that a polynomial of degree $n$ has exactly $n$ complex roots (zeros), counting multiplicity. So, $R(x)$ has exactly 10 zeros in total. We are given that $4-2i$ is a zero. By the Complex Conjugate Root Theorem, we know $4+2i$ is also a zero. These are two non-real complex zeros. We are also given that $-2$ is a zero with a multiplicity of 3. This means that the factor $(x+2)$ appears three times, contributing three real zeros (since -2 is a real number). So far, we have identified:

• $4-2i$ (1 complex zero)
• $4+2i$ (1 complex zero)
• $-2$ (with multiplicity 3, meaning 3 real zeros)

In total, we have accounted for $1 + 1 + 3 = 5$ zeros. Since the polynomial has a degree of 10, there are $10 - 5 = 5$ remaining zeros.

These remaining 5 zeros can be either real or complex. However, we know that any additional complex zeros must also come in conjugate pairs because the coefficients are real. If we want to find the maximum number of real zeros, we should assume that all of the remaining 5 zeros are real numbers. This assumption maximizes the count of real roots.

So, the total number of real zeros would be:

• The 3 real zeros from the multiplicity of $-2$.
• The 5 additional real zeros we are assuming to maximize the count.

Adding these together, $3 + 5 = 8$. Therefore, the maximum number of real zeros $R(x)$ can have is 8. The other two zeros must be the complex conjugate pair $4-2i$ and $4+2i$. This distribution satisfies the degree requirement (8 real + 2 complex = 10 total zeros) and the conditions given in the problem.

It's important to contrast this with the minimum number of real zeros. To minimize real zeros, we would maximize complex conjugate pairs among the remaining 5 zeros. We could have two pairs (4 complex zeros) and one real zero. This would give $3 + 1 = 4$ real zeros. The number of real zeros can therefore range from 4 to 8. The question specifically asks for the maximum, which is 8.

This problem highlights a fundamental concept in algebra: the relationship between the degree of a polynomial, its coefficients, and the nature of its zeros. The constraint of having real coefficients is key, as it forces complex zeros to appear in symmetric conjugate pairs. Understanding these principles allows us to deduce properties of polynomials even when we don't have all the information explicitly laid out. It's a testament to the logical structure and predictive power inherent in algebra.

For further reading on polynomial properties and theorems, you can explore resources like Wolfram MathWorld or Khan Academy.