Finding Factors Of Polynomial Functions With Roots 3+√5 And -6

Hey guys! Ever wondered how to find the factors of a polynomial function when you're given roots that involve radicals? It might sound intimidating, but it's actually a pretty cool concept once you get the hang of it. Let's dive into a problem where we're given a polynomial function, ${ f(x) }$, with roots ${ 3 + \sqrt{5} }$ and -6, and we need to figure out a factor of ${ f(x) }$. This is a classic algebra problem that touches on some fundamental ideas about polynomials and their roots. So, grab your thinking caps, and let's get started!

Understanding Polynomial Roots and Factors

First things first, let's quickly recap what roots and factors actually are in the context of polynomials. The roots of a polynomial function are the values of ${ x }$ that make the function equal to zero. In other words, if ${ r }$ is a root of ${ f(x) }$, then ${ f(r) = 0 }$. Factors, on the other hand, are expressions that divide evenly into the polynomial. There's a direct connection between roots and factors: if ${ r }$ is a root of ${ f(x) }$, then ${ (x - r) }$ is a factor of ${ f(x) }$. This is a cornerstone concept in polynomial algebra, and it's what allows us to build polynomials from their roots and vice versa. So, when we're given a root, we immediately know a corresponding factor, which is super handy for solving problems like this one.

Now, let's think about what happens when a root is an irrational number, like ${ 3 + \sqrt{5} }$. Irrational numbers can't be expressed as a simple fraction, and they often come with a twist when they appear as roots of polynomials with rational coefficients. This twist is the conjugate root theorem. This theorem is our secret weapon when dealing with irrational roots, especially those involving square roots. The conjugate root theorem states that if a polynomial with rational coefficients has a root of the form ${ a + \sqrt{b} }$, where ${ a }$ and ${ b }$ are rational and ${ \sqrt{b} }$ is irrational, then its conjugate ${ a - \sqrt{b} }$ must also be a root. This is because irrational roots involving square roots often arise from quadratic factors with rational coefficients, and these factors always have conjugate pairs as roots. So, if ${ 3 + \sqrt{5} }$ is a root, we immediately know that its conjugate, ${ 3 - \sqrt{5} }$, is also a root. This is a crucial piece of information for finding factors of the polynomial.

Applying the Conjugate Root Theorem

In our case, we know that ${ f(x) }$ has a root of ${ 3 + \sqrt{5} }$. Thanks to the conjugate root theorem, we also know that ${ 3 - \sqrt{5} }$ must be a root. This is awesome because it gives us two roots to work with! Remember, each root corresponds to a factor. So, if ${ 3 + \sqrt{5} }$ is a root, then ${ (x - (3 + \sqrt{5})) }$ is a factor. Similarly, since ${ 3 - \sqrt{5} }$ is a root, then ${ (x - (3 - \sqrt{5})) }$ is a factor. But wait, there's more! We also know that -6 is a root, which means ${ (x - (-6)) }$, or simply ${ (x + 6) }$, is another factor. This is like finding puzzle pieces that fit together to reveal the bigger picture of the polynomial.

Now, let's zoom in on the options given in the problem. We're looking for a factor of ${ f(x) }$. Option A is ${ (x + (3 - \sqrt{5})) }$, which simplifies to ${ (x + 3 - \sqrt{5}) }$. This doesn't directly match any of the factors we've identified so far. Option B is ${ (x - (3 - \sqrt{5})) }$. Aha! This matches one of the factors we found using the conjugate root theorem. This is a strong contender. Option C, ${ (x + (5 + \sqrt{3})) }$, doesn't seem to be related to our roots at all. So, just by applying the conjugate root theorem and the basic relationship between roots and factors, we're already honing in on the correct answer. The key here is to methodically apply the theorems and definitions we know to break down the problem into manageable parts. Let's take a closer look at why option B is indeed the correct answer.

Why Option B is the Correct Factor

Option B, ${ (x - (3 - \sqrt{5})) }$, is the correct factor because it directly corresponds to the root ${ 3 - \sqrt{5} }$, which we know is a root of ${ f(x) }$ due to the conjugate root theorem. Remember, the conjugate root theorem is our superpower in this situation. It allows us to instantly identify another root when we're given an irrational root involving a square root. This theorem is not just a random fact; it's deeply connected to the structure of polynomials with rational coefficients. When we have a quadratic factor that produces an irrational root like ${ 3 + \sqrt{5} }$, the conjugate ${ 3 - \sqrt{5} }$ is automatically generated as the other root. This is because the quadratic formula, which is used to solve for the roots of a quadratic equation, involves a ${ \pm }$ sign before the square root term. This ${ \pm }$ sign ensures that if one root has the form ${ a + \sqrt{b} }$, the other root will have the form ${ a - \sqrt{b} }$.

Let's break down how ${ (x - (3 - \sqrt{5})) }$ becomes a factor. When we have a root ${ r }$, the corresponding factor is always ${ (x - r) }$. In this case, our root ${ r }$ is ${ 3 - \sqrt{5} }$. So, we substitute this into ${ (x - r) }$ to get ${ (x - (3 - \sqrt{5})) }$. This is exactly what option B gives us! We can even take it a step further and simplify this factor. Distributing the negative sign, we get ${ x - 3 + \sqrt{5} }$. This expression, ${ x - 3 + \sqrt{5} }$, is a factor of ${ f(x) }$. It's important to note that while this factor might look a bit complex with the irrational term, it's still a perfectly valid factor of the polynomial. It's these kinds of factors that give polynomials their interesting properties and behaviors. So, option B is not just a possible answer; it's the definitive, correct factor based on the roots we were given and the conjugate root theorem. This is why understanding the underlying principles and theorems is so crucial in math. They give us the tools to tackle even seemingly complex problems with confidence.

Constructing the Polynomial (Optional)

Just for kicks, let's think about how we could actually construct a polynomial ${ f(x) }$ that has these roots. We know the roots are ${ 3 + \sqrt{5} }$, ${ 3 - \sqrt{5} }$, and -6. Each root gives us a factor. The factors are ${ (x - (3 + \sqrt{5})) }$, ${ (x - (3 - \sqrt{5})) }$, and ${ (x + 6) }$. To get a polynomial, we multiply these factors together. Let's start by multiplying the factors corresponding to the conjugate roots:

${ (x - (3 + \sqrt{5}))(x - (3 - \sqrt{5})) }$

This might look messy, but let's simplify it. We can rewrite this as:

${ ((x - 3) - \sqrt{5})((x - 3) + \sqrt{5}) }$

Notice that this is in the form ${ (a - b)(a + b) }$, which we know expands to ${ a^2 - b^2 }$. So, we have:

${ (x - 3)^2 - (\sqrt{5})^2 }$

Expanding this, we get:

${ (x^2 - 6x + 9) - 5 }$

Which simplifies to:

${ x^2 - 6x + 4 }$

Now, we multiply this quadratic factor by the factor corresponding to the root -6, which is ${ (x + 6) }$:

${ (x^2 - 6x + 4)(x + 6) }$

Expanding this gives us:

${ x^3 + 6x^2 - 6x^2 - 36x + 4x + 24 }$

Which simplifies to:

${ x^3 - 32x + 24 }$

So, one possible polynomial ${ f(x) }$ with the given roots is ${ x^3 - 32x + 24 }$. This is just one example; we could multiply this polynomial by any constant and it would still have the same roots. The key takeaway here is that knowing the roots allows us to construct the polynomial, and understanding the relationship between roots and factors is fundamental to this process. This step isn't necessary to answer the original question, but it's a fantastic way to deepen our understanding of the concepts involved. It shows how the roots, factors, and polynomial all fit together in a beautiful, interconnected way.

Conclusion: Mastering Polynomial Factors

So, to wrap it up, when we're given a polynomial function ${ f(x) }$ with roots ${ 3 + \sqrt{5} }$ and -6, the conjugate root theorem tells us that ${ 3 - \sqrt{5} }$ is also a root. This means that ${ (x - (3 - \sqrt{5})) }$ is a factor of ${ f(x) }$, making option B the correct answer. Guys, remember that understanding the relationship between roots and factors, along with key theorems like the conjugate root theorem, is super important for tackling these kinds of problems. These concepts aren't just abstract ideas; they're powerful tools that allow us to break down and solve complex problems in algebra and beyond. Keep practicing, and you'll become a polynomial pro in no time!

By understanding these principles, you'll be well-equipped to tackle similar problems and gain a deeper understanding of the fascinating world of polynomials. Keep practicing, and you'll master these concepts in no time!