Polynomial Function With Complex Roots A Step-by-Step Solution

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Hey guys! Ever found yourself scratching your head over polynomial functions, especially when those pesky imaginary roots pop up? You're definitely not alone! In this guide, we're going to break down a common type of polynomial problem: finding a polynomial function with a leading coefficient of 1 and specific complex roots. We'll tackle the concepts, the steps, and even look at a sample problem to make sure you've got it down pat. So, buckle up and let's dive into the fascinating world of polynomials!

Understanding the Basics: Roots and Polynomials

Before we jump into the nitty-gritty, let's make sure we're all on the same page with the fundamental concepts. The roots of a polynomial function, also known as zeros or solutions, are the values of x that make the function equal to zero. In simpler terms, they're the points where the graph of the polynomial intersects the x-axis (if we're dealing with real roots).

Polynomials are mathematical expressions consisting of variables and coefficients, combined using addition, subtraction, and non-negative integer exponents. A polynomial function is a function defined by a polynomial expression. For instance, f(x) = x² + 3x - 4 is a polynomial function.

A crucial concept here is the Fundamental Theorem of Algebra. This theorem states that a polynomial of degree n (the highest power of x) has exactly n complex roots, counting multiplicities. What does this mean? Well, it tells us that every polynomial equation has solutions, and it also gives us an idea of how many solutions to expect. For example, a polynomial of degree 2 (a quadratic) will have two roots, a polynomial of degree 3 (a cubic) will have three roots, and so on.

Now, let's talk about complex roots. These are roots that involve the imaginary unit i, where i is defined as the square root of -1. Complex roots always come in conjugate pairs. This means that if a + bi is a root of a polynomial with real coefficients, then a - bi must also be a root. This is a super important point to remember when constructing polynomials with complex roots.

Finally, the leading coefficient is the coefficient of the term with the highest power of x in the polynomial. It plays a significant role in determining the end behavior of the polynomial's graph. A leading coefficient of 1 simply means that the coefficient of the highest degree term is 1.

Constructing Polynomials from Roots: The Key Steps

Okay, now that we've got the basics covered, let's move on to the main event: constructing a polynomial function from its roots. Here's the general recipe:

  1. Identify all the roots, including their multiplicities: Remember that complex roots come in conjugate pairs. If you're given a complex root like a + bi, you automatically know that a - bi is also a root. The multiplicity of a root tells you how many times that root appears as a solution. For instance, a root with multiplicity 2 means it's a repeated root.
  2. Write the factors: For each root r, write a factor of the form (x - r). If a root r has a multiplicity of m, the corresponding factor will be (x - r)ᵐ.
  3. Multiply the factors: Multiply all the factors together to obtain the polynomial. This might involve some algebraic manipulation, like expanding products of binomials.
  4. Adjust the leading coefficient (if necessary): If the problem specifies a leading coefficient other than 1, multiply the entire polynomial by the appropriate constant.

Let's break down these steps with a simple example. Suppose we want to find a polynomial with roots 2 and -3. Following the steps:

  1. Identify roots: The roots are 2 and -3.
  2. Write factors: The factors are (x - 2) and (x + 3).
  3. Multiply factors: (x - 2)(x + 3) = x² + x - 6
  4. Adjust leading coefficient: If we want a leading coefficient of 1, we're already good to go! The polynomial is f(x) = x² + x - 6.

Dealing with Complex Roots: A Closer Look

Things get a little more interesting when we throw complex roots into the mix. But don't worry, the core principles remain the same. The key thing to remember is that complex roots come in conjugate pairs. Let's walk through an example to illustrate this.

Suppose we want to find a polynomial with roots 2 + i and 2 - i. Notice that these are conjugates of each other, just as we'd expect. Here's how we construct the polynomial:

  1. Identify roots: The roots are 2 + i and 2 - i.

  2. Write factors: The factors are (x - (2 + i)) and (x - (2 - i)).

  3. Multiply factors: This is where things get a little more involved. Let's multiply these factors carefully:

    (x - (2 + i))(x - (2 - i)) = (x - 2 - i)(x - 2 + i)

    Now, we can use the distributive property (or the FOIL method) to expand this product:

    = x² - 2x + ix - 2x + 4 - 2i - ix + 2i - i²

    Notice that the terms ix and -ix cancel out, and the terms -2i and 2i also cancel out. Also, remember that i² = -1. So we have:

    = x² - 4x + 4 - (-1)

    = x² - 4x + 5

  4. Adjust leading coefficient: Again, if we want a leading coefficient of 1, we're set. The polynomial is f(x) = x² - 4x + 5.

See how the imaginary terms disappeared when we multiplied the factors corresponding to the conjugate roots? This is a general pattern, and it's why complex roots always come in conjugate pairs when we're dealing with polynomials with real coefficients.

Solving the Sample Problem: A Step-by-Step Approach

Alright, let's bring it all together and tackle the problem presented: Which polynomial function has a leading coefficient of 1 and roots 2i and 3i with multiplicity 1?

Here's how we'll break it down:

  1. Identify all the roots: We're given the roots 2i and 3i. Since these are complex roots, we know their conjugates, -2i and -3i, must also be roots. Each root has a multiplicity of 1, meaning they each appear once.

  2. Write the factors: The factors corresponding to these roots are (x - 2i), (x + 2i), (x - 3i), and (x + 3i).

  3. Multiply the factors: Let's multiply these factors together. It's often easier to multiply conjugate pairs first:

    (x - 2i)(x + 2i) = x² - (2i)² = x² - (-4) = x² + 4

    (x - 3i)(x + 3i) = x² - (3i)² = x² - (-9) = x² + 9

    Now, multiply these two results:

    (x² + 4)(x² + 9) = x⁴ + 9x² + 4x² + 36 = x⁴ + 13x² + 36

  4. Adjust the leading coefficient: The leading coefficient of our resulting polynomial is already 1, so we don't need to do any further adjustments.

Therefore, the polynomial function with a leading coefficient of 1 and roots 2i and 3i (with multiplicity 1) is f(x) = x⁴ + 13x² + 36. Notice that this is a quartic polynomial (degree 4), which makes sense since we have four roots (two given and their two conjugates).

Looking at the options provided, we can see that none of them directly match our result. However, option A, f(x) = (x - 2i)(x - 3i), only considers the given roots and doesn't account for their conjugates. Option B, f(x) = (x + 2i)(x + 3i), also has the same issue. Option C, f(x) = (x - 2)(x - 3)(x - 2i)(x - 3i), includes real roots (2 and 3) that were not specified in the problem.

Therefore, none of the provided options are correct. The correct polynomial function is f(x) = x⁴ + 13x² + 36.

Key Takeaways and Tips for Success

  • Complex roots always come in conjugate pairs: This is a fundamental concept when dealing with polynomials with real coefficients.
  • Multiply conjugate pairs first: This simplifies the multiplication process and eliminates imaginary terms.
  • Pay attention to multiplicities: A root with multiplicity m corresponds to a factor raised to the power of m.
  • Check the degree of the polynomial: The degree should match the total number of roots (counting multiplicities).
  • Don't be afraid to expand and simplify:** Algebraic manipulation is often necessary to obtain the final polynomial.
  • Practice, practice, practice: The more you work through these types of problems, the more comfortable you'll become with the process.

By understanding the relationship between roots and factors, and by remembering the key concepts like the Fundamental Theorem of Algebra and the conjugate pairs theorem, you'll be well-equipped to tackle any polynomial problem that comes your way. Keep practicing, and you'll become a polynomial pro in no time! Remember, mastering these concepts not only helps in academics but also provides a solid foundation for advanced studies in mathematics, engineering, and other related fields. So, keep up the great work, guys!

Further Exploration and Practice Problems

To solidify your understanding, try working through some additional practice problems. Here are a few ideas:

  1. Find a polynomial function with a leading coefficient of 1 and roots 1 + i, 1 - i, and 3.
  2. Find a polynomial function with a leading coefficient of 2 and roots i (with multiplicity 2) and -i (with multiplicity 2).
  3. Find a polynomial function with roots -2, 0, and 3 + 2i.

By working through these problems, you'll develop a deeper understanding of the concepts and hone your problem-solving skills. Don't hesitate to refer back to this guide as needed, and remember that persistence is key to success in mathematics. Happy solving, and may the polynomials be ever in your favor! Remember guys, the journey of learning is a marathon, not a sprint. Take your time, enjoy the process, and celebrate each milestone you achieve. You've got this!