Is (x-3) A Factor? Best Justification For Polynomial P(x)

Hey guys! Let's dive into a super important concept in polynomial algebra: determining factors of polynomials. Today, we're tackling the question of how to best justify whether (x-3) is a factor of the polynomial p(x) = x^3 - 3x^2 - 2x - 6. This is a classic problem that pops up in algebra courses, and understanding the underlying principles will seriously level up your math game. So, buckle up, and let's break it down!

Understanding the Factor Theorem

At the heart of this problem is the Factor Theorem. This theorem is your best friend when you're trying to figure out if a linear expression (like x - a) is a factor of a polynomial. In simple terms, the Factor Theorem states:

A polynomial p(x) has a factor (x - a) if and only if p(a) = 0.

Let's unpack that. What it means is if you substitute a into the polynomial p(x) and the result is zero, then (x - a) is definitely a factor of p(x). Conversely, if (x - a) is a factor of p(x), then plugging in a will make the polynomial equal to zero. This "if and only if" relationship is super powerful. Think of it like a two-way street: it works in both directions. To really nail this down, let’s look at why this works.

When we say (x - a) is a factor of p(x), we're saying that p(x) can be written as (x - a) times some other polynomial, let's call it q(x). So, p(x) = (x - a) * q(x). Now, if we substitute x = a into this equation, we get:

p(a) = (a - a) * q(a) = 0 * q(a) = 0

See? It all falls into place. The (a - a) part becomes zero, which makes the whole thing zero. This is the magic of the Factor Theorem in action. This principle is key to efficiently solving these types of problems. Instead of going through long division or other cumbersome methods, you can simply substitute a value and check if the result is zero. If it is, you’ve got a factor! If not, move on. This approach saves time and reduces the chance of making mistakes.

Why the Remainder Matters

Now, let's talk about the remainder. When you divide a polynomial p(x) by (x - a), you'll get a quotient q(x) and a remainder r. This relationship can be written as:

p(x) = (x - a) * q(x) + r

The remainder r is crucial. If r is 0, it means that (x - a) divides p(x) evenly, with no leftover. This is exactly what we want when we're looking for factors. If r is anything other than 0, then (x - a) is not a factor of p(x). So, the remainder is like a direct signal telling you whether you've found a factor or not. When the remainder is zero, it’s like a green light signaling that (x-a) is indeed a factor. A non-zero remainder, on the other hand, is a red light, telling you to look elsewhere.

Applying the Factor Theorem to Our Problem

Okay, with the Factor Theorem fresh in our minds, let’s get back to our specific problem: Is (x - 3) a factor of p(x) = x^3 - 3x^2 - 2x - 6? To find out, we'll use the Factor Theorem. We need to substitute x = 3 (because we're checking for the factor x - 3) into the polynomial p(x) and see what we get.

Let’s do it:

p(3) = (3)^3 - 3(3)^2 - 2(3) - 6 p(3) = 27 - 3(9) - 6 - 6 p(3) = 27 - 27 - 6 - 6 p(3) = -12

So, p(3) = -12. What does this tell us? Well, according to the Factor Theorem, since p(3) is not 0, then (x - 3) is not a factor of p(x). It's that simple! We've used the Factor Theorem to directly answer our question. The result, -12, immediately indicates that (x-3) doesn’t divide the polynomial evenly. If p(3) had equaled zero, we would have had a factor, but in this case, we know it's not.

Why the Other Options Aren't the Best Justifications

Now, let’s consider why the other options might not be the best justifications. Often, in math problems, there are multiple true statements, but one is more precise or directly answers the question. Let's think about it.

If we look at options claiming that (x - 3) is a factor, we know those are incorrect because we've already calculated p(3) and found it to be -12, not 0. These options might mention remainders or dividing evenly, but they don't match our calculation. It's crucial to directly apply the Factor Theorem and see the result for yourself.

Conclusion: The Best Justification

So, what’s the best justification? Based on the Factor Theorem and our calculations, the correct justification is:

It is not a factor of p(x) because the remainder is not 0.

This statement gets straight to the point. It highlights the core concept – the remainder – and connects it directly to the Factor Theorem. When the remainder isn't zero, the linear expression isn't a factor. Period. Remember, the beauty of the Factor Theorem is its efficiency and clarity. It gives us a direct way to check for factors without needing to perform long division or other lengthy calculations. By substituting the value into the polynomial and checking the result, we can quickly determine whether we have a factor or not. This approach not only saves time but also provides a solid understanding of the underlying mathematical principles.

I hope this explanation helps you guys understand how to use the Factor Theorem to determine if a linear expression is a factor of a polynomial. Keep practicing, and you'll become a polynomial pro in no time! This stuff might seem tricky at first, but with practice, you'll be able to tackle these problems with confidence. Remember, math is like building a puzzle; each piece you learn helps you see the bigger picture. So, keep learning, keep practicing, and keep having fun with math!