Finding Values Of A And B When Polynomial Has Specific Factors
Introduction
In this comprehensive article, we will delve into the problem of determining the values of the constants a and b such that the cubic polynomial p(x) = 2x³ + ax² + x + b has the linear expressions (x + 2) and (2x - 1) as its factors. This problem elegantly combines concepts from polynomial factorization, the factor theorem, and solving systems of equations. Understanding these concepts is crucial for success in algebra and calculus, and this exploration will serve as a valuable exercise for students and enthusiasts alike.
Problem Statement
Our core task is to find the values of a and b that make (x + 2) and (2x - 1) factors of the polynomial p(x) = 2x³ + ax² + x + b. This means that when p(x) is divided by either (x + 2) or (2x - 1), the remainder should be zero. We will leverage the factor theorem, which provides a direct link between the roots of a polynomial and its factors, to solve this problem effectively.
Key Concepts and Theorems
Before diving into the solution, let's solidify our understanding of the key concepts involved:
- Polynomial Factorization: Polynomial factorization involves expressing a polynomial as a product of simpler polynomials. In our case, we aim to express p(x) as a product involving (x + 2) and (2x - 1).
- Factor Theorem: The factor theorem states that for a polynomial p(x), (x - c) is a factor if and only if p(c) = 0. This theorem is the cornerstone of our solution strategy, as it allows us to translate the factor condition into algebraic equations.
- Roots of a Polynomial: The roots of a polynomial p(x) are the values of x for which p(x) = 0. These roots correspond to the x-intercepts of the polynomial's graph.
- System of Equations: As we apply the factor theorem, we will obtain a system of linear equations in a and b. Solving this system will give us the desired values of a and b.
Applying the Factor Theorem
To effectively find the values of a and b, let's employ the factor theorem. Since (x + 2) is a factor of p(x), then p(-2) = 0. Substituting x = -2 into p(x) gives us:
p(-2) = 2(-2)³ + a(-2)² + (-2) + b = 0
Simplifying this equation, we get:
-16 + 4a - 2 + b = 0
4a + b = 18 (Equation 1)
Similarly, since (2x - 1) is a factor of p(x), then p(1/2) = 0. Substituting x = 1/2 into p(x) gives us:
p(1/2) = 2(1/2)³ + a(1/2)² + (1/2) + b = 0
Simplifying this equation, we get:
1/4 + a/4 + 1/2 + b = 0
Multiplying through by 4 to eliminate fractions, we obtain:
1 + a + 2 + 4b = 0
a + 4b = -3 (Equation 2)
Solving the System of Equations
Now we have a system of two linear equations in two variables, a and b:
4a + b = 18 (Equation 1)
a + 4b = -3 (Equation 2)
We can solve this system using several methods, such as substitution or elimination. Let's use the elimination method.
Multiply Equation 2 by -4 to eliminate a:
-4(a + 4b) = -4(-3)
-4a - 16b = 12 (Equation 3)
Now add Equation 1 and Equation 3:
(4a + b) + (-4a - 16b) = 18 + 12
-15b = 30
Divide by -15:
b = -2
Now that we have the value of b, we can substitute it back into either Equation 1 or Equation 2 to find the value of a. Let's use Equation 2:
a + 4(-2) = -3
a - 8 = -3
a = 5
Therefore, the values of a and b are a = 5 and b = -2.
Verification
To ensure the correctness of our solution, we should verify that with a = 5 and b = -2, both (x + 2) and (2x - 1) are indeed factors of p(x) = 2x³ + 5x² + x - 2. Let's perform polynomial division.
First, we'll divide p(x) by (x + 2):
2x² + x - 1
x + 2 | 2x³ + 5x² + x - 2
- (2x³ + 4x²)
----------------
x² + x
- (x² + 2x)
-------------
-x - 2
- (-x - 2)
-----------
0
This division confirms that (x + 2) is a factor, and we obtain the quotient 2x² + x - 1. Now, let's check if (2x - 1) is a factor of this quotient.
The quotient 2x² + x - 1 can be factored as (2x - 1)(x + 1). This confirms that (2x - 1) is indeed a factor of the quotient and, consequently, a factor of the original polynomial p(x).
Therefore, our values a = 5 and b = -2 are correct.
Alternative Method: Polynomial Long Division
An alternative method to find the values of a and b is to perform polynomial long division directly. We can divide 2x³ + ax² + x + b by (x + 2)(2x - 1) and set the remainder to zero.
First, expand (x + 2)(2x - 1):
(x + 2)(2x - 1) = 2x² - x + 4x - 2 = 2x² + 3x - 2
Now, perform the long division of 2x³ + ax² + x + b by 2x² + 3x - 2:
x + (a-3)/2
2x² + 3x - 2 | 2x³ + ax² + x + b
- (2x³ + 3x² - 2x)
------------------------
(a-3)x² + 3x + b
- ((a-3)x² + 3(a-3)/2 x - (a-3))
----------------------------------
(3 - 3(a-3)/2)x + b + a - 3
For (2x² + 3x - 2) to be a factor, the remainder must be zero. This gives us two equations:
3 - 3(a - 3)/2 = 0
b + a - 3 = 0
Solving the first equation:
3 = 3(a - 3)/2
2 = a - 3
a = 5
Substituting a = 5 into the second equation:
b + 5 - 3 = 0
b + 2 = 0
b = -2
This method also confirms that a = 5 and b = -2.
Conclusion
In conclusion, we have successfully found the values of a and b that make (x + 2) and (2x - 1) factors of the polynomial p(x) = 2x³ + ax² + x + b. Through the application of the factor theorem and solving the resulting system of linear equations, we determined that a = 5 and b = -2. We verified our solution using polynomial long division and provided an alternative method involving direct polynomial division by the product of the factors. This exercise highlights the interconnectedness of algebraic concepts and provides a solid foundation for more advanced problem-solving in mathematics.
This exploration underscores the importance of understanding and applying fundamental theorems like the factor theorem. The ability to manipulate polynomials, solve equations, and verify solutions is crucial for success in mathematics and related fields. This problem serves as a valuable learning experience, reinforcing key concepts and enhancing problem-solving skills.
