Find Polynomials: Leading Coeff 3, Roots -4, I, 2

Hey math whizzes! Today we're diving deep into the world of polynomial functions, and trust me, guys, it's more fun than it sounds. We've got a specific challenge: find the polynomial function that has a leading coefficient of 3 and includes the roots -4, i, and 2. Oh, and here's the kicker – all these roots have a multiplicity of 1. This means each root appears exactly once in the factored form of the polynomial. When we talk about the leading coefficient, we're referring to the coefficient of the term with the highest power of x in the polynomial. It's like the superstar of the polynomial, setting the overall scale and direction. The roots, on the other hand, are the x-values where the function equals zero, essentially where the graph of the polynomial crosses the x-axis. Understanding these properties is key to reconstructing the polynomial from its building blocks.

Let's break down how we can construct such a polynomial. Remember the Factor Theorem? It's a super handy tool that tells us if a value 'a' is a root of a polynomial, then (x - a) must be a factor of that polynomial. So, if we know our roots are -4, i, and 2, we can immediately start building our factors: (x - (-4)), (x - i), and (x - 2). Simplifying these, we get (x + 4), (x - i), and (x - 2).

Now, here's where things get a little tricky but also super interesting. When dealing with polynomial functions, especially those with real coefficients (which is usually implied unless stated otherwise), complex roots always come in conjugate pairs. This means if 'i' (which is 0 + 1i) is a root, then its conjugate, '-i' (which is 0 - 1i), must also be a root. So, even though the problem only explicitly mentioned 'i' as a root, the fact that we're likely working with a polynomial with real coefficients means we must include '-i' as a root as well. This adds another factor to our list: (x - (-i)), which simplifies to (x + i).

So, our complete set of factors, considering both the given roots and the necessary conjugate pair, is (x + 4), (x - i), (x + i), and (x - 2). Now, we need to assemble these factors into a polynomial. The general form of a polynomial with given roots and a leading coefficient is $f(x) = a(x - r_1)(x - r_2)...(x - r_n)$, where 'a' is the leading coefficient and $r_1, r_2, ..., r_n$ are the roots. In our case, the leading coefficient 'a' is given as 3.

Putting it all together, the polynomial function should look something like this: $f(x) = 3(x + 4)(x - i)(x + i)(x - 2)$.

Let's simplify the complex factors first because multiplying conjugates is a neat trick. Remember that $(a - b)(a + b) = a^2 - b^2$? Here, 'a' is 'x' and 'b' is 'i'. So, $(x - i)(x + i) = x^2 - i^2$. Since $i^2 = -1$, this becomes $x^2 - (-1) = x^2 + 1$.

Now, substitute this back into our polynomial expression: $f(x) = 3(x + 4)(x^2 + 1)(x - 2)$.

This is the factored form of our polynomial. The question provides multiple-choice options, and we need to identify which one matches our derived function. Let's examine the options:

• Option A: $f(x)=3(x+4)(x-1)(x-2)$. This option has the correct leading coefficient and the roots -4 and 2, but it includes the root 1 instead of 'i' and '-i'. Also, the root 1 has a multiplicity of 1, which is correct for the roots it lists, but it doesn't account for the complex root 'i'. This is incorrect.
• Option B: $f(x)=(x-3)(x+4)(x-1)(x-2)$. This option has the roots -4 and 2, but the leading coefficient is 1 (not 3), and it has roots 3 and 1, which are not the required complex roots. This is incorrect.
• Option C: $f(x)=(x-3)(x+4)(x-1)(x+1)(x-2)$. This option has roots 3, -4, 1, -1, and 2. The leading coefficient is 1, and it doesn't include the required complex root 'i'. This is incorrect.
• Option D: $f(x)=3(x+4)(x-1)(x+1)(x-2)$. This option has a leading coefficient of 3 and includes the roots -4 and 2. However, it lists the roots 1 and -1 instead of the complex roots 'i' and '-i'. This is incorrect.

Wait a minute, guys! It seems there might be a slight misunderstanding or a typo in the provided options when compared to the exact requirements of the problem. The problem states the roots are -4, i, and 2. As we established, for a polynomial with real coefficients, 'i' must be accompanied by its conjugate '-i'. This gives us factors $(x+4)$, $(x-i)$, $(x+i)$, and $(x-2)$. The leading coefficient is 3. So, the correct function should be $f(x) = 3(x+4)(x^2+1)(x-2)$.

Let's re-evaluate the options with this understanding. It seems the options might be testing a slightly different scenario or there's a common mistake pattern they are trying to present. Often, when 'i' is mentioned, students might incorrectly assume it's a typo for '1' or that only the explicitly stated roots matter. However, the rule about complex conjugate pairs is fundamental for polynomials with real coefficients.

Let's assume, for the sake of argument, that the question intended to provide options that did include the complex roots, or perhaps there's a way the options are phrased that we need to decipher. If we look closely at the structure, options A and D have the correct leading coefficient of 3 and the root -4 and 2. The difference lies in the other factors. Option A has $(x-1)$ and Option D has $(x-1)(x+1)$.

Now, let's reconsider the problem statement very carefully.