Polynomial Factorization And Remainder Theorem Determining Coefficients And Complete Factorization

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Introduction

In this comprehensive exploration, we delve into the fascinating world of polynomials, focusing on the intricate relationships between factors, the Remainder Theorem, and the determination of unknown coefficients. Polynomials, fundamental building blocks in algebra, often present intriguing challenges that require a deep understanding of their properties and behavior. This article aims to dissect a specific problem involving a cubic polynomial with unknown coefficients, leveraging the concepts of polynomial factorization and the Remainder Theorem to unveil the values of these elusive constants. Understanding the relationship between factors and polynomials is a crucial concept in algebra. A factor of a polynomial is an expression that divides evenly into the polynomial, leaving no remainder. In other words, if a polynomial P(x) has a factor (x - a), then P(a) = 0. This is a direct consequence of the Factor Theorem, which is a special case of the Remainder Theorem.

Problem Statement

We are given that the expression $(x^2 - 4)$ is a factor of the cubic polynomial $P(x) = ax^3 + bx^2 - 4x + c$. Furthermore, we are told that when this polynomial is divided by $(x + 1)$, the remainder is 6. Our mission is twofold:

(a) Determine the values of the unknown coefficients a, b, and c.

(b) (The problem statement for part (b) is missing, but we will explore a possible extension related to factoring the polynomial completely.)

This problem beautifully intertwines the concepts of factorization and the Remainder Theorem. The fact that $(x^2 - 4)$ is a factor provides us with valuable information about the roots of the polynomial, while the remainder condition gives us another equation to work with. By strategically applying these concepts, we can unravel the values of the coefficients and gain a deeper understanding of the polynomial's structure. The Remainder Theorem provides a powerful shortcut for finding the remainder when a polynomial is divided by a linear expression. It states that if a polynomial P(x) is divided by (x - a), then the remainder is P(a). This theorem is particularly useful when we are given the remainder and need to find the value of the polynomial at a specific point. The combination of factorization and the Remainder Theorem allows us to solve a wide range of problems involving polynomials, including those with unknown coefficients.

Part (a): Finding the Values of a, b, and c

Step 1: Utilizing the Factor $(x^2 - 4)$

Since $(x^2 - 4)$ is a factor of $P(x)$, we can express $P(x)$ as a product of $(x^2 - 4)$ and another linear factor. Notice that $(x^2 - 4)$ can be further factored as a difference of squares: $(x^2 - 4) = (x - 2)(x + 2)$. This tells us that x = 2 and x = -2 are roots of the polynomial $P(x)$. Therefore, $P(2) = 0$ and $P(-2) = 0$. This is a direct application of the Factor Theorem, which states that if x = a is a root of a polynomial P(x), then (x - a) is a factor of P(x), and vice-versa. By identifying the roots from the given factor, we can substitute these values into the polynomial and obtain equations that relate the unknown coefficients. This is a powerful technique for simplifying the problem and making it more manageable.

Step 2: Applying P(2) = 0

Substituting x = 2 into the polynomial $P(x) = ax^3 + bx^2 - 4x + c$, we get:

P(2)=a(2)3+b(2)2−4(2)+c=8a+4b−8+c=0P(2) = a(2)^3 + b(2)^2 - 4(2) + c = 8a + 4b - 8 + c = 0

This gives us our first equation:

8a+4b+c=88a + 4b + c = 8

This equation establishes a linear relationship between the coefficients a, b, and c. It represents a plane in the three-dimensional space of (a, b, c), and any point on this plane satisfies the condition that x = 2 is a root of the polynomial. The next step will involve finding another such equation, which will allow us to narrow down the possible solutions for the coefficients. This process of substituting known roots into the polynomial is a fundamental technique in polynomial algebra, and it is crucial for solving problems involving unknown coefficients.

Step 3: Applying P(-2) = 0

Similarly, substituting x = -2 into $P(x)$, we obtain:

P(−2)=a(−2)3+b(−2)2−4(−2)+c=−8a+4b+8+c=0P(-2) = a(-2)^3 + b(-2)^2 - 4(-2) + c = -8a + 4b + 8 + c = 0

This yields our second equation:

−8a+4b+c=−8-8a + 4b + c = -8

Now we have two equations involving a, b, and c. These equations represent two planes in the three-dimensional space of (a, b, c), and the intersection of these planes will give us a line of possible solutions for the coefficients. The next step will involve using the Remainder Theorem to obtain a third equation, which will allow us to pinpoint a unique solution for a, b, and c. The ability to translate polynomial conditions into algebraic equations is a key skill in problem-solving, and this example demonstrates how the Factor Theorem can be used to generate such equations.

Step 4: Utilizing the Remainder Theorem

We are given that when $P(x)$ is divided by $(x + 1)$, the remainder is 6. By the Remainder Theorem, this means that $P(-1) = 6$. Substituting x = -1 into $P(x)$, we get:

P(−1)=a(−1)3+b(−1)2−4(−1)+c=−a+b+4+c=6P(-1) = a(-1)^3 + b(-1)^2 - 4(-1) + c = -a + b + 4 + c = 6

This gives us our third equation:

−a+b+c=2-a + b + c = 2

We now have a system of three linear equations with three unknowns:

  1. 8a+4b+c=88a + 4b + c = 8

  2. −8a+4b+c=−8-8a + 4b + c = -8

  3. −a+b+c=2-a + b + c = 2

This system of equations can be solved using various methods, such as substitution, elimination, or matrix methods. The goal is to find a unique set of values for a, b, and c that satisfy all three equations simultaneously. The Remainder Theorem has provided us with the crucial third equation needed to solve this system, highlighting the importance of this theorem in polynomial algebra.

Step 5: Solving the System of Equations

Let's solve the system of equations. Adding equations (1) and (2), we eliminate a:

(8a+4b+c)+(−8a+4b+c)=8+(−8)(8a + 4b + c) + (-8a + 4b + c) = 8 + (-8)

8b+2c=08b + 2c = 0

4b+c=0ext(Equation4)4b + c = 0 ext{ (Equation 4)}

Now, subtract equation (3) from equation (2):

(−8a+4b+c)−(−a+b+c)=−8−2(-8a + 4b + c) - (-a + b + c) = -8 - 2

−7a+3b=−10ext(Equation5)-7a + 3b = -10 ext{ (Equation 5)}

From Equation 4, we have $c = -4b$. Substitute this into equation (3):

−a+b+(−4b)=2-a + b + (-4b) = 2

−a−3b=2-a - 3b = 2

a=−3b−2ext(Equation6)a = -3b - 2 ext{ (Equation 6)}

Substitute Equation 6 into Equation 5:

−7(−3b−2)+3b=−10-7(-3b - 2) + 3b = -10

21b+14+3b=−1021b + 14 + 3b = -10

24b=−2424b = -24

b=−1b = -1

Now, substitute b = -1 into Equation 6:

a=−3(−1)−2=3−2=1a = -3(-1) - 2 = 3 - 2 = 1

Finally, substitute b = -1 into $c = -4b$:

c=−4(−1)=4c = -4(-1) = 4

Therefore, we have found the values of the coefficients: a = 1, b = -1, and c = 4. This solution satisfies all three equations in the system, confirming that it is the correct set of coefficients for the polynomial. The process of solving a system of linear equations is a fundamental skill in mathematics, and this example demonstrates its application in the context of polynomial algebra.

Step 6: Verification

Let's verify our solution by substituting the values a = 1, b = -1, and c = 4 back into the original polynomial and the given conditions.

The polynomial becomes:

P(x)=x3−x2−4x+4P(x) = x^3 - x^2 - 4x + 4

We know that $(x^2 - 4) = (x - 2)(x + 2)$ should be a factor. Let's perform polynomial long division or synthetic division to check if $P(x)$ is divisible by $(x^2 - 4)$:

Dividing $P(x)$ by $(x^2 - 4)$, we get:

(x3−x2−4x+4)/(x2−4)=x−1(x^3 - x^2 - 4x + 4) / (x^2 - 4) = x - 1

Since the division is exact (no remainder), $(x^2 - 4)$ is indeed a factor of $P(x)$.

Now, let's check the remainder when $P(x)$ is divided by $(x + 1)$:

P(−1)=(−1)3−(−1)2−4(−1)+4=−1−1+4+4=6P(-1) = (-1)^3 - (-1)^2 - 4(-1) + 4 = -1 - 1 + 4 + 4 = 6

This confirms that the remainder is 6, as given in the problem statement.

Therefore, our solution a = 1, b = -1, and c = 4 is correct. The verification step is a crucial part of the problem-solving process, as it ensures that the solution obtained satisfies all the given conditions. This step helps to catch any potential errors made during the calculations and provides confidence in the final answer.

Part (b): Factoring the Polynomial Completely

Now that we have found the coefficients, let's proceed to factor the polynomial completely. We know that $P(x) = x^3 - x^2 - 4x + 4$ and that $(x^2 - 4)$ is a factor. From our verification in Part (a), we found that:

P(x)=(x2−4)(x−1)P(x) = (x^2 - 4)(x - 1)

We can further factor $(x^2 - 4)$ as a difference of squares:

x2−4=(x−2)(x+2)x^2 - 4 = (x - 2)(x + 2)

Therefore, the complete factorization of $P(x)$ is:

P(x)=(x−2)(x+2)(x−1)P(x) = (x - 2)(x + 2)(x - 1)

This complete factorization reveals the three roots of the polynomial: x = 2, x = -2, and x = 1. The roots are the values of x for which the polynomial is equal to zero. The complete factorization provides a comprehensive understanding of the polynomial's behavior and its relationship to the x-axis. It allows us to easily identify the points where the polynomial intersects the x-axis, which are the roots of the polynomial.

Significance of Complete Factorization

The complete factorization of a polynomial is a fundamental concept in algebra and has significant implications in various areas of mathematics and its applications. Some of the key benefits of complete factorization include:

  • Finding Roots: As demonstrated above, the roots of the polynomial are directly revealed by the complete factorization. These roots are crucial for solving polynomial equations and understanding the behavior of the polynomial function.
  • Graphing Polynomials: The roots, along with the leading coefficient and the degree of the polynomial, provide valuable information for sketching the graph of the polynomial function. The roots indicate the x-intercepts, and the leading coefficient and degree determine the end behavior of the graph.
  • Solving Inequalities: Complete factorization is essential for solving polynomial inequalities. By identifying the intervals where the polynomial is positive or negative, we can determine the solution set for the inequality.
  • Simplifying Rational Expressions: Factoring both the numerator and denominator of a rational expression allows us to simplify the expression by canceling common factors. This simplification is crucial for performing algebraic operations on rational expressions.
  • Calculus Applications: In calculus, factorization is used in various contexts, such as finding the critical points of a function, determining the intervals of increasing and decreasing behavior, and evaluating limits.

The ability to factor polynomials completely is a fundamental skill in algebra, and it is essential for success in higher-level mathematics courses.

Conclusion

This problem has provided us with a rich exploration of polynomial factorization, the Remainder Theorem, and the process of solving for unknown coefficients. By strategically applying these concepts, we successfully determined the values of a, b, and c and factored the polynomial completely. Mastering these techniques is crucial for tackling more complex problems in algebra and beyond. The problem also highlighted the importance of verification in the problem-solving process, ensuring that the solution obtained satisfies all the given conditions. The complete factorization of the polynomial revealed its roots and provided a deeper understanding of its behavior. The concepts and techniques discussed in this article are fundamental to polynomial algebra and have wide-ranging applications in mathematics and related fields. From finding roots to solving equations and inequalities, the ability to manipulate and understand polynomials is a cornerstone of mathematical proficiency. The journey through this problem has not only provided a solution but also reinforced our understanding of these core algebraic principles. Moving forward, these skills will serve as a solid foundation for tackling more advanced mathematical challenges and exploring the beauty and power of polynomial algebra. Understanding the relationship between the coefficients and roots of a polynomial is a cornerstone of algebraic knowledge. The ability to connect these concepts is vital for problem-solving and for building a solid foundation in mathematics. In this exploration, we have seen how the Factor Theorem and Remainder Theorem provide powerful tools for unraveling the mysteries of polynomials and their behavior. As we continue our mathematical journey, we will undoubtedly encounter more challenging and fascinating problems involving polynomials. The key to success lies in a deep understanding of the fundamental concepts and the ability to apply them creatively and strategically. The world of polynomials is vast and intricate, but with persistence and a thirst for knowledge, we can unlock its secrets and appreciate its elegance and power.