Polynomial With Integer Coefficients: Find P(x)
Let's dive into how to find a polynomial p(x) that not only has integer coefficients but also satisfies specific conditions. We're on the hunt for a polynomial of degree 4 with zeros at x = 3 and x = 1/2. Plus, we want the one with the smallest positive leading coefficient. Sounds like a fun challenge, right? Let's break it down step by step.
Understanding the Basics of Polynomials
Before we jump into solving this specific problem, let's make sure we're all on the same page with the basics of polynomials. Polynomials are algebraic expressions that consist of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents. They're the building blocks of many equations and functions in mathematics, playing a crucial role in everything from calculus to cryptography. When we talk about the degree of a polynomial, we're referring to the highest power of the variable in the polynomial. So, a polynomial of degree 4 will have a term with x raised to the power of 4.
A zero of a polynomial, also known as a root, is a value of x that makes the polynomial equal to zero. In other words, if p(a) = 0, then a is a zero of the polynomial p(x). Zeros are incredibly important because they tell us where the polynomial crosses or touches the x-axis on a graph. Understanding the relationship between zeros and factors is key to constructing polynomials. If a is a zero of p(x), then (x - a) is a factor of p(x). This is a fundamental concept that we'll use extensively in our quest to find p(x). The leading coefficient is the coefficient of the term with the highest power of x. This coefficient plays a crucial role in determining the overall shape and behavior of the polynomial, especially as x gets very large or very small. In our case, we're looking for the polynomial with the smallest positive leading coefficient, which adds an extra layer of optimization to the problem.
Constructing the Polynomial
Now, let's get our hands dirty and start constructing the polynomial p(x). We know that x = 3 and x = 1/2 are zeros of our polynomial. This means that (x - 3) and (x - 1/2) are factors of p(x). But wait, we need integer coefficients, right? Having a fractional zero like 1/2 might seem like a problem, but we can easily fix it. Instead of (x - 1/2), we can use (2x - 1) as a factor. This ensures that when x = 1/2, the factor becomes zero, and we've cleverly eliminated the fraction. Since our polynomial p(x) is of degree 4, we need two more factors. To keep things simple and find the polynomial with the smallest positive leading coefficient, let's consider the possibility of repeated roots or simple linear factors. The simplest approach is to consider the factors (x - 3) and (2x - 1) occurring twice. So, we can express our polynomial p(x) in the form:
p(x) = a(x - 3)(x - 3)(2x - 1)(2x - 1)
Here, a is the leading coefficient that we want to minimize while keeping it positive. Expanding this expression will give us a polynomial of degree 4, and we can then ensure that all coefficients are integers. This form directly incorporates the given zeros and allows us to adjust the leading coefficient to meet our criteria. Let's expand this expression and see what we get.
Expanding and Simplifying
Okay, let's expand and simplify the expression p(x) = a(x - 3)(x - 3)(2x - 1)(2x - 1). First, let's multiply (x - 3)(x - 3), which is simply (x - 3)^2. This gives us:
(x - 3)^2 = x^2 - 6x + 9
Next, let's multiply (2x - 1)(2x - 1), which is (2x - 1)^2. This results in:
(2x - 1)^2 = 4x^2 - 4x + 1
Now, we need to multiply these two quadratic expressions together:
(x^2 - 6x + 9)(4x^2 - 4x + 1)
This might look a bit intimidating, but we'll take it term by term. Multiplying each term in the first expression by each term in the second expression, we get:
4x^4 - 4x^3 + x^2 - 24x^3 + 24x^2 - 6x + 36x^2 - 36x + 9
Now, let's combine like terms:
4x^4 + (-4x^3 - 24x^3) + (x^2 + 24x^2 + 36x^2) + (-6x - 36x) + 9
Simplifying this gives us:
4x^4 - 28x^3 + 61x^2 - 42x + 9
So, our polynomial p(x) now looks like:
p(x) = a(4x^4 - 28x^3 + 61x^2 - 42x + 9)
Remember, we want the smallest positive leading coefficient a. Since the leading coefficient inside the parenthesis is 4, the smallest positive integer value for a is 1. This ensures that the leading coefficient of the entire polynomial remains as small as possible. Therefore, we can set a = 1. This simplifies our polynomial to:
p(x) = 4x^4 - 28x^3 + 61x^2 - 42x + 9
This is a polynomial of degree 4 with integer coefficients and zeros at x = 3 and x = 1/2, and it has the smallest positive leading coefficient. Nice job, guys! We've successfully constructed the polynomial we were looking for.
Verifying the Solution
Alright, before we pat ourselves on the back, let's make sure our solution is correct. Verification is a crucial step in problem-solving, especially in mathematics. We've found the polynomial p(x) = 4x^4 - 28x^3 + 61x^2 - 42x + 9, and we need to verify that it meets all the given conditions. First, let's check if x = 3 is a zero of p(x). We substitute x = 3 into the polynomial:
p(3) = 4(3)^4 - 28(3)^3 + 61(3)^2 - 42(3) + 9
p(3) = 4(81) - 28(27) + 61(9) - 42(3) + 9
p(3) = 324 - 756 + 549 - 126 + 9
p(3) = 0
Great! x = 3 is indeed a zero of our polynomial. Now, let's check if x = 1/2 is also a zero:
p(1/2) = 4(1/2)^4 - 28(1/2)^3 + 61(1/2)^2 - 42(1/2) + 9
p(1/2) = 4(1/16) - 28(1/8) + 61(1/4) - 42(1/2) + 9
p(1/2) = 1/4 - 7/2 + 61/4 - 21 + 9
To simplify this, let's find a common denominator, which is 4:
p(1/2) = 1/4 - 14/4 + 61/4 - 84/4 + 36/4
p(1/2) = (1 - 14 + 61 - 84 + 36) / 4
p(1/2) = 0 / 4
p(1/2) = 0
Excellent! x = 1/2 is also a zero of p(x). We've confirmed that our polynomial has the required zeros. Next, let's confirm that it's a polynomial of degree 4. Looking at our expression, p(x) = 4x^4 - 28x^3 + 61x^2 - 42x + 9, the highest power of x is 4, so it's indeed a degree 4 polynomial. Finally, we need to ensure that the leading coefficient is the smallest positive integer. The leading coefficient of our polynomial is 4, which is positive. We chose a = 1 to minimize this coefficient, so we've met this condition as well. After thorough verification, we can confidently say that our solution, p(x) = 4x^4 - 28x^3 + 61x^2 - 42x + 9, is correct. We've successfully found a polynomial with integer coefficients, degree 4, zeros at x = 3 and x = 1/2, and the smallest positive leading coefficient. Great job, team!
Alternative Approaches and Considerations
While we've successfully found our polynomial p(x), it's always a good idea to consider alternative approaches and delve a bit deeper into the problem. This not only solidifies our understanding but also equips us with different problem-solving tools for future challenges. One alternative approach could involve exploring different combinations of factors. We chose to repeat the factors (x - 3) and (2x - 1), but what if we had chosen different factors? For instance, we could have considered quadratic factors with integer coefficients that have no real roots. This would lead to a different polynomial that still satisfies the given conditions, but it might not have the smallest positive leading coefficient. Another consideration is the uniqueness of the solution. Is the polynomial we found the only polynomial that satisfies the given conditions with the smallest positive leading coefficient? To explore this, we could think about the general form of a degree 4 polynomial with the given zeros:
p(x) = a(x - 3)(2x - 1)(bx^2 + cx + d)
Here, a, b, c, and d are integers, and the quadratic factor (bx^2 + cx + d) has no real roots. By expanding this and analyzing the coefficients, we could potentially find other polynomials. However, ensuring that the leading coefficient is the smallest positive integer and that all coefficients are integers adds constraints that limit the possibilities. Furthermore, we could think about the graphical representation of the polynomial. A polynomial of degree 4 can have at most 4 real roots. Since we already have two real roots (x = 3 and x = 1/2), the other two roots could be real or complex. If they are complex, they must occur in conjugate pairs, which means the quadratic factor (bx^2 + cx + d) would have complex roots. Exploring these alternative approaches and considerations enhances our understanding of polynomials and problem-solving strategies. It also highlights the importance of not just finding a solution but also understanding the solution's context and uniqueness. Keep up the great work, guys! You're doing awesome!
Conclusion
So, we've successfully navigated the challenge of finding a polynomial p(x) with integer coefficients, a degree of 4, zeros at x = 3 and x = 1/2, and the smallest positive leading coefficient. We started by understanding the basics of polynomials, zeros, and factors. Then, we constructed the polynomial using the given zeros, expanded and simplified the expression, and verified our solution. We even explored alternative approaches and considerations to deepen our understanding. The polynomial we found is p(x) = 4x^4 - 28x^3 + 61x^2 - 42x + 9. This exercise not only reinforces our knowledge of polynomials but also showcases the power of systematic problem-solving. By breaking down a complex problem into smaller, manageable steps, we can tackle even the trickiest mathematical challenges. Remember, the key is to understand the fundamental concepts, apply them methodically, and always verify your solution. And don't forget to explore different perspectives and approaches – it's what makes problem-solving truly rewarding. Keep practicing, keep exploring, and keep pushing your mathematical boundaries. You've got this! Hitting the books and enjoying the process, that's what it is all about. Mathematical adventures await, guys!