Find Polynomial Constant Term: Remainder Theorem

h1. Finding the Polynomial's Constant Term Using the Remainder Theorem

Have you ever encountered a math problem where you're given a polynomial, told that dividing it by a certain expression leaves no remainder, and then asked to find a missing part of the polynomial? This is a classic scenario where the Remainder Theorem comes to your rescue! In this article, we'll dive deep into how to solve problems like: "When the polynomial $3x^3 - 5x^2 - 47x + k$ is divided by $(x+3)$, the remainder is 0. What is the value of the polynomial's constant term, k?" We'll break down the concept, explore the underlying theorem, and walk through the solution step-by-step, making it crystal clear how to tackle these types of questions. Understanding this concept is crucial for mastering polynomial algebra and is a common topic in algebra courses.

h2. Understanding the Remainder Theorem

The Remainder Theorem is a fundamental concept in algebra that provides a shortcut for finding the remainder of a polynomial division. Stated simply, it says that if a polynomial $P(x)$ is divided by a linear expression $(x - a)$, then the remainder is equal to $P(a)$. This theorem is incredibly powerful because it allows us to find the remainder without actually performing the long division. Think about it: instead of going through the whole process of dividing polynomials, which can be tedious and error-prone, we can just substitute a single value into the polynomial. It's like having a secret key that unlocks the answer directly! This theorem is a direct consequence of the polynomial division algorithm, which states that for any polynomial $P(x)$ and any non-zero polynomial $D(x)$, there exist unique polynomials $Q(x)$ (the quotient) and $R(x)$ (the remainder) such that $P(x) = D(x) Q(x) + R(x)$, where the degree of $R(x)$ is less than the degree of $D(x)$. When the divisor $D(x)$ is a linear term of the form $(x-a)$, the degree of the remainder $R(x)$ must be less than 1, meaning $R(x)$ is a constant, let's call it $R$. So, $P(x) = (x-a) Q(x) + R$. If we substitute $x = a$ into this equation, we get $P(a) = (a-a) Q(a) + R$, which simplifies to $P(a) = 0 imes Q(a) + R$, and thus, $P(a) = R$. This is precisely what the Remainder Theorem states. It's elegant, efficient, and a must-know for any aspiring mathematician or student grappling with polynomial functions. The beauty of this theorem lies in its simplicity and its broad applicability in various algebraic problems.

h3. Applying the Remainder Theorem to Our Problem

Let's get back to our specific problem: "When the polynomial $P(x) = 3x^3 - 5x^2 - 47x + k$ is divided by $(x+3)$, the remainder is 0. What is the value of the polynomial's constant term, k?" Here, our polynomial is $P(x) = 3x^3 - 5x^2 - 47x + k$. The divisor is $(x+3)$. According to the Remainder Theorem, if $P(x)$ is divided by $(x-a)$, the remainder is $P(a)$. In our case, the divisor is $(x+3)$, which can be written as $(x - (-3))$. Therefore, $a = -3$. The problem states that the remainder is 0. This means that $P(-3)$ must equal 0. So, our task is to substitute $-3$ for $x$ in the polynomial $P(x)$ and set the entire expression equal to 0.

This gives us the equation:

$P(-3) = 3(-3)^3 - 5(-3)^2 - 47(-3) + k = 0$

Now, we need to carefully evaluate each term. Remember the order of operations (PEMDAS/BODMAS): Parentheses/Brackets, Exponents/Orders, Multiplication and Division (from left to right), Addition and Subtraction (from left to right). First, let's handle the exponents:

$(-3)^3 = (-3) imes (-3) imes (-3) = 9 imes (-3) = -27$

$(-3)^2 = (-3) imes (-3) = 9$

Now, substitute these values back into the equation:

$3(-27) - 5(9) - 47(-3) + k = 0$

Next, perform the multiplications:

$3 imes (-27) = -81$

$-5 imes 9 = -45$

$-47 imes (-3) = 141$

Substitute these results:

$-81 - 45 + 141 + k = 0$

Now, combine the constant terms:

$-81 - 45 = -126$

$-126 + 141 = 15$

So, the equation simplifies to:

$15 + k = 0$

To find the value of k, we simply subtract 15 from both sides of the equation:

$k = -15$

And there you have it! The value of the polynomial's constant term, k, is -15. This entire process beautifully illustrates the power and efficiency of the Remainder Theorem in solving algebraic problems involving polynomials. It's a concept that, once understood, can significantly simplify complex calculations and deepen your appreciation for the elegance of mathematics.

h2. Why the Remainder is Zero: The Factor Theorem

In our problem, the remainder was given as 0. This is a special case that leads us to another important theorem: the Factor Theorem. The Factor Theorem is actually a direct corollary of the Remainder Theorem. It states that for a polynomial $P(x)$, $(x-a)$ is a factor of $P(x)$ if and only if $P(a) = 0$. In simpler terms, if dividing a polynomial $P(x)$ by $(x-a)$ results in a remainder of 0, it means that $(x-a)$ divides the polynomial evenly, and therefore, $(x-a)$ is a factor of that polynomial. In our specific problem, since the remainder is 0 when $P(x) = 3x^3 - 5x^2 - 47x + k$ is divided by $(x+3)$, it means that $(x+3)$ is a factor of the polynomial. This also means that $x = -3$ is a root (or a zero) of the polynomial equation $P(x) = 0$. This concept is foundational for understanding polynomial roots and factorization, which are key components in solving higher-level algebraic equations and functions. When $(x-a)$ is a factor, it implies that when you substitute $x=a$ into the polynomial, the result is zero. This is because $P(x) = (x-a)Q(x)$ when $(x-a)$ is a factor, and thus $P(a) = (a-a)Q(a) = 0 imes Q(a) = 0$. This relationship is incredibly useful when you need to find the roots of a polynomial or determine if a given linear expression is a factor. It provides a direct link between factors, roots, and the value of the polynomial at specific points. The elegance of these theorems lies in their ability to connect different aspects of polynomial behavior in a unified and comprehensible way. Understanding the Factor Theorem alongside the Remainder Theorem equips you with a robust toolkit for analyzing and manipulating polynomials effectively.

h3. Verifying Our Answer

Let's quickly verify our answer to ensure that $k=-15$ is indeed correct. With $k=-15$, our polynomial becomes $P(x) = 3x^3 - 5x^2 - 47x - 15$. According to the Remainder Theorem, when this polynomial is divided by $(x+3)$, the remainder should be $P(-3)$. Let's calculate $P(-3)$:

$P(-3) = 3(-3)^3 - 5(-3)^2 - 47(-3) - 15$

$P(-3) = 3(-27) - 5(9) - 47(-3) - 15$

$P(-3) = -81 - 45 + 141 - 15$

$P(-3) = -126 + 141 - 15$

$P(-3) = 15 - 15$

$P(-3) = 0$

As expected, the remainder is 0. This confirms that our calculated value of $k=-15$ is correct. The fact that the remainder is 0 also implies, by the Factor Theorem, that $(x+3)$ is a factor of the polynomial $3x^3 - 5x^2 - 47x - 15$. This is a powerful confirmation that our application of the Remainder Theorem was accurate and that our value for the constant term is correct. It's always a good practice to double-check your work, especially in mathematics, as it solidifies your understanding and helps catch any potential errors. This verification step not only proves the correctness of our solution but also reinforces the underlying mathematical principles at play, making the learning experience more robust and memorable. The consistency between the Remainder Theorem and the Factor Theorem provides a solid foundation for confidence in the derived answer.

h2. Conclusion

We've successfully navigated the process of finding the constant term of a polynomial using the Remainder Theorem. The key takeaway is that when a polynomial $P(x)$ is divided by $(x-a)$ and the remainder is $R$, then $P(a) = R$. In our specific problem, the remainder was 0, meaning $P(-3) = 0$. By substituting $x = -3$ into the polynomial $3x^3 - 5x^2 - 47x + k$ and setting it equal to 0, we were able to solve for the unknown constant term $k$, finding it to be -15. This problem elegantly demonstrates how the Remainder Theorem simplifies complex calculations and provides direct insights into polynomial behavior. Understanding these theorems is not just about solving homework problems; it's about building a strong foundation in algebraic manipulation and problem-solving skills that are transferable to many areas of mathematics and beyond. Keep practicing these concepts, and you'll find yourself more comfortable and confident tackling even more challenging polynomial problems.

For further exploration and practice on polynomial theorems, you can visit Khan Academy's comprehensive resources on algebra. They offer detailed explanations, examples, and exercises that can further enhance your understanding of the Remainder Theorem and its applications. Additionally, exploring MathWorld by Wolfram Research provides in-depth mathematical definitions and theorems, including those related to polynomials.