Finding Polynomial H(x) With Complex Zero & Given Values
Hey guys! Let's dive into a fascinating problem involving polynomial functions. We're given a polynomial function, h(x), that has a zero at x = 2 - 5i with a multiplicity of one. That’s a fancy way of saying that one of the roots of the polynomial is the complex number 2 - 5i. Remember, complex roots always come in conjugate pairs, which is a crucial concept we'll use later. We also have a table of values for h(x) at certain points. Our mission, should we choose to accept it, is to figure out as much as we can about this polynomial. This means understanding its structure, its roots, and how the given values help us nail down its specific form. Polynomial functions are the workhorses of algebra and calculus, appearing in countless applications, from modeling physical phenomena to designing algorithms. So, grasping the intricacies of their behavior, especially when complex roots are involved, is a super valuable skill. We'll break down the problem step by step, making sure to highlight the key concepts and techniques. So buckle up, and let's get started on this polynomial adventure! We are going to learn how to piece together the puzzle of a polynomial function, using complex roots and given values as our guide.
Understanding the Complex Zero
Okay, first things first: the problem tells us that h(x) has a zero at x = 2 - 5i. What does this actually mean? Well, a zero of a function is simply a value of x that makes the function equal to zero. In other words, if we plug 2 - 5i into h(x), we get 0 as the result. But here's the really important part: because the coefficients of our polynomial h(x) are real numbers (we're assuming this, as it's standard for these types of problems), complex roots always come in conjugate pairs. This is a fundamental theorem in algebra. What's a conjugate pair, you ask? It’s simply a complex number with the sign of its imaginary part flipped. So, the conjugate of 2 - 5i is 2 + 5i. This means that if 2 - 5i is a zero, then 2 + 5i must also be a zero of h(x). This is awesome news! We’ve just doubled our knowledge of the roots of the polynomial. Knowing these complex roots is like having a secret key to unlocking the structure of h(x). Each root corresponds to a factor of the polynomial. For example, if x = a is a root, then (x - a) is a factor. So, the complex roots 2 - 5i and 2 + 5i give us two factors that we can use to start building our polynomial. This is a powerful technique that allows us to move from knowing the roots to actually constructing the polynomial itself. The conjugate pair theorem is not just some abstract mathematical concept; it’s a practical tool that helps us solve problems involving polynomials with complex roots. Remember this, guys: complex roots travel in pairs!
Constructing Factors from Zeros
Now that we know we have two zeros, x = 2 - 5i and x = 2 + 5i, let's translate this knowledge into factors of our polynomial h(x). Remember, a zero x = a corresponds to a factor of (x - a). So, for the zero x = 2 - 5i, we get the factor (x - (2 - 5i)), which simplifies to (x - 2 + 5i). Similarly, for the zero x = 2 + 5i, we get the factor (x - (2 + 5i)), which simplifies to (x - 2 - 5i). Great! We have two factors. But let’s make things a bit cleaner. We can multiply these two complex factors together to get a quadratic factor with real coefficients. This is a clever trick that avoids dealing with complex numbers directly in the polynomial. When we multiply (x - 2 + 5i) and (x - 2 - 5i), we're essentially multiplying two complex conjugates. A cool thing happens when you do this: the imaginary terms cancel out. Let's do the multiplication:
(x - 2 + 5i)(x - 2 - 5i) = (x - 2)^2 - (5i)^2.
Expanding this, we get:
(x^2 - 4x + 4) - (-25) = x^2 - 4x + 29.
Ta-da! We have a quadratic factor x^2 - 4x + 29, which corresponds to the two complex roots. Notice that this quadratic has real coefficients, which is exactly what we wanted. This is a super useful step because it allows us to work with a simpler expression while still capturing the information about the complex roots. This quadratic factor is now part of the puzzle that makes up our polynomial h(x). But we’re not done yet! We still need to consider the other information given in the problem, specifically the table of values.
Incorporating the Real Zero
Alright, let's shift our focus. Besides the complex zero, we also have a table of values for h(x). The table tells us that h(-5) = 0. Boom! This is huge. This means that x = -5 is also a zero of the polynomial. This is a real zero, which is different from the complex zeros we dealt with earlier. Since x = -5 is a zero, we know that (x - (-5)), which simplifies to (x + 5), must be another factor of h(x). Now we're cooking with gas! We've got a quadratic factor from the complex roots and a linear factor from the real root. We're starting to see the shape of our polynomial take form. Remember, the degree of a polynomial (the highest power of x) tells us the maximum number of roots it can have. Since we have a quadratic factor (degree 2) and a linear factor (degree 1), our polynomial must have at least degree 3. This means it could be a cubic polynomial (degree 3), or it could have a higher degree if there are other roots we don't know about yet. But for now, let's focus on what we do know. We can combine the factors we've found so far to write a partial expression for h(x). We know that h(x) must be a multiple of both (x^2 - 4x + 29) and (x + 5). So, we can write h(x) = a(x^2 - 4x + 29)(x + 5), where a is some constant. We need to figure out what this constant a is. This is where the remaining values from the table come into play. These values act like clues, helping us solve for the missing piece of the puzzle.
Using Table Values to Determine the Leading Coefficient
Okay, we've made some serious progress! We know that h(x) = a(x^2 - 4x + 29)(x + 5), where a is a constant we need to find. This is where the rest of the table values come to the rescue. The table gives us specific points on the graph of h(x), and we can use these points to solve for a. Let's pick one of the points from the table. How about h(-2) = 3? This means that when we plug in x = -2 into our expression for h(x), we should get 3 as the result. Let's do it:
3 = a((-2)^2 - 4(-2) + 29)(-2 + 5)
Now we just need to simplify and solve for a. Let's break it down:
3 = a(4 + 8 + 29)(3)
3 = a(41)(3)
3 = 123a
Now, divide both sides by 123:
a = 3/123 = 1/41
Fantastic! We've found the value of a. It's 1/41. This means we now have the complete expression for our polynomial h(x). We can substitute this value back into our equation:
h(x) = (1/41)(x^2 - 4x + 29)(x + 5)
This is our polynomial! We started with a complex zero and a few table values, and we've successfully constructed the entire polynomial function. This is a testament to the power of using the conjugate pair theorem, factoring, and leveraging given values to solve for unknowns. But hold on, we're not quite finished. It's always a good idea to double-check our work. Let’s use another point from the table to verify that our polynomial is correct.
Verifying the Polynomial
Alright, we've got our candidate for h(x): h(x) = (1/41)(x^2 - 4x + 29)(x + 5). But before we declare victory, let's make absolutely sure it's the real deal. The best way to do this is to test it with another point from the table that we haven't used yet. Let's pick h(-1) = 2. This means if we plug in x = -1 into our polynomial, we should get 2 as the output. Let's see if it holds true:
2 = (1/41)((-1)^2 - 4(-1) + 29)(-1 + 5)
Now we simplify, step by step:
2 = (1/41)(1 + 4 + 29)(4)
2 = (1/41)(34)(4)
2 = (1/41)(136)
2 = 136/41
Oops! It seems like something's not quite right. 136/41 is not equal to 2. This means there might be a mistake in our calculations somewhere, or perhaps there's something else we need to consider about the polynomial. Don't worry, this is a normal part of the problem-solving process. Sometimes things don't work out perfectly the first time, and that's okay. It just means we need to go back and carefully review our steps to see where we might have gone wrong. This is where the real learning happens! Let’s take a deep breath and retrace our steps, checking each calculation and assumption we made along the way. It's possible we made a small arithmetic error, or maybe we need to think more deeply about the structure of the polynomial. Let's start by reviewing our calculations for the constant a. It's a good idea to double-check this, as it's a crucial part of our solution.
Identifying and Correcting the Error
Okay, guys, let's put on our detective hats and retrace our steps. Remember when we calculated the constant a using h(-2) = 3? We had the equation:
3 = a((-2)^2 - 4(-2) + 29)(-2 + 5)
Let's break this down again and make sure we didn't make any sneaky arithmetic errors:
3 = a(4 + 8 + 29)(3)
3 = a(41)(3)
3 = 123a
So far, so good. Now we divide both sides by 123:
a = 3/123 = 1/41
Okay, the calculation for a still seems correct. So, the issue isn't there. Hmmm… what else could it be? Let's go back to our polynomial expression:
h(x) = (1/41)(x^2 - 4x + 29)(x + 5)
And remember when we tested it with h(-1) = 2? We got:
2 = (1/41)((-1)^2 - 4(-1) + 29)(-1 + 5)
2 = (1/41)(1 + 4 + 29)(4)
2 = (1/41)(34)(4)
2 = (1/41)(136)
2 = 136/41
This is where we realized something was off. But wait a minute… let's look really closely at the original problem. Did we make any assumptions that might not be valid? Aha! The problem only states that the zero x = 2 - 5i has a multiplicity of one. It doesn't say that x = -5 has a multiplicity of one. This is a crucial distinction! What if x = -5 has a multiplicity greater than one? This would mean that (x + 5) appears as a factor multiple times. Let's explore this possibility. If (x + 5) has a multiplicity of 2, then we'd have a factor of (x + 5)^2. Our polynomial would then look like:
h(x) = a(x^2 - 4x + 29)(x + 5)^2
This is a game-changer! Let's see if this new form of h(x) can explain the values in the table.
Refining the Polynomial with Multiplicity
Okay, we've had a major breakthrough! We realized that the zero x = -5 might have a multiplicity greater than one. This led us to consider the possibility that our polynomial has the form:
h(x) = a(x^2 - 4x + 29)(x + 5)^2
where a is still a constant we need to determine. Let's use the value h(-2) = 3 again, but this time with our new polynomial form. Plugging in x = -2, we get:
3 = a((-2)^2 - 4(-2) + 29)(-2 + 5)^2
Now, let's simplify:
3 = a(4 + 8 + 29)(3)^2
3 = a(41)(9)
3 = 369a
Now, solve for a:
a = 3/369 = 1/123
Great! We have a new value for a: 1/123. So, our updated polynomial is:
h(x) = (1/123)(x^2 - 4x + 29)(x + 5)^2
This looks promising! But before we get too excited, let's verify this polynomial with the other table value we have: h(-1) = 2. Plugging in x = -1, we get:
2 = (1/123)((-1)^2 - 4(-1) + 29)(-1 + 5)^2
Let's simplify:
2 = (1/123)(1 + 4 + 29)(4)^2
2 = (1/123)(34)(16)
2 = (1/123)(544)
2 = 544/123
Hold on a second… This still doesn't quite equal 2. We're getting closer, but it's not a perfect match. This suggests that even our updated polynomial might not be the complete picture. We might need to consider an even higher multiplicity for the root x = -5, or perhaps there's another factor we're missing altogether. The problem-solving process can be like this sometimes – it’s a journey of exploration and refinement. We've learned a lot along the way, even though we haven't arrived at the final answer just yet. Let's take a step back again and think about what other possibilities might exist.
Exploring Higher Multiplicities and Other Factors
Okay, we've been wrestling with this polynomial for a while now, and we've learned some valuable lessons. We initially assumed a multiplicity of one for the real root x = -5, but that didn't quite fit the data. Then we tried a multiplicity of two, and we got closer, but still not a perfect match. So, what's next? Should we try a multiplicity of three for x = -5? That's definitely a possibility. If we did that, our polynomial would look like:
h(x) = a(x^2 - 4x + 29)(x + 5)^3
But before we jump to that, let's think about the degree of the polynomial. If we have a quadratic factor and a cubic factor, that would give us a polynomial of degree 5. While this is mathematically possible, it might be more complex than what's intended for this type of problem. It's always a good idea to look for the simplest solution first. So, let's consider another possibility: what if there's another real root that we haven't accounted for yet? Remember, polynomials can have multiple roots. We know about the complex conjugate pair and the root x = -5, but there could be another one lurking in the shadows. If there's another real root, let's call it x = r. This would give us another factor of (x - r) in our polynomial. Our polynomial would then look like:
h(x) = a(x^2 - 4x + 29)(x + 5)(x - r)
This form of the polynomial has degree 4, which is a reasonable degree for this type of problem. Now we have two unknowns: the constant a and the root r. We'll need two pieces of information to solve for these unknowns. Luckily, we have two table values we haven't fully used yet: h(-2) = 3 and h(-1) = 2. This is promising! We can plug these values into our polynomial and get two equations with two unknowns. Then we can use algebra to solve for a and r. This might seem like a bit of a daunting task, but we've come this far, and we have the tools we need to crack this problem. Let's set up those equations and see where they lead us.
Setting Up a System of Equations
Alright, we're on the hunt for a new real root, r, and we're going to use the table values h(-2) = 3 and h(-1) = 2 to help us find it. Our polynomial now has the form:
h(x) = a(x^2 - 4x + 29)(x + 5)(x - r)
Let's plug in x = -2 and h(-2) = 3:
3 = a((-2)^2 - 4(-2) + 29)(-2 + 5)(-2 - r)
Simplifying, we get:
3 = a(4 + 8 + 29)(3)(-2 - r)
3 = a(41)(3)(-2 - r)
3 = 123a(-2 - r)
Now, let's plug in x = -1 and h(-1) = 2:
2 = a((-1)^2 - 4(-1) + 29)(-1 + 5)(-1 - r)
Simplifying, we get:
2 = a(1 + 4 + 29)(4)(-1 - r)
2 = a(34)(4)(-1 - r)
2 = 136a(-1 - r)
We now have two equations:
- 3 = 123a(-2 - r)
- 2 = 136a(-1 - r)
This is a system of two equations with two unknowns, a and r. We can use various methods to solve this system, such as substitution or elimination. Let's use elimination. First, let's divide the first equation by 123 and the second equation by 136:
- 3/123 = a(-2 - r) => 1/41 = a(-2 - r)
- 2/136 = a(-1 - r) => 1/68 = a(-1 - r)
Now we have:
- 1/41 = -2a - ar
- 1/68 = -a - ar
Let's subtract the second equation from the first equation. This will eliminate the ar term:
(1/41) - (1/68) = -2a - ar - (-a - ar)
(68 - 41) / (41 * 68) = -2a + a
27 / 2788 = -a
So,
a = -27 / 2788
Now that we have a, we can plug it back into either equation to solve for r. Let's use the second equation:
1/68 = (-27 / 2788)(-1 - r)
Multiply both sides by 68:
1 = (-27 * 68 / 2788)(-1 - r)
1 = (-27 / 41)(-1 - r)
Multiply both sides by -41/27:
-41/27 = -1 - r
Add 1 to both sides:
-41/27 + 1 = -r
(-41 + 27) / 27 = -r
-14/27 = -r
So,
r = 14/27
We've done it! We've found the values of a and r. Now we can write the complete polynomial:
h(x) = (-27 / 2788)(x^2 - 4x + 29)(x + 5)(x - 14/27)
This was a challenging problem, but we persevered, and we arrived at a solution. We used a combination of factoring, the conjugate pair theorem, and a system of equations to find the polynomial function h(x). Give yourself a pat on the back for sticking with it!
Final Answer and Reflections
Alright, guys! After a journey filled with twists, turns, and a bit of detective work, we've finally arrived at our polynomial function:
h(x) = (-27 / 2788)(x^2 - 4x + 29)(x + 5)(x - 14/27)
This is the polynomial that satisfies all the conditions given in the problem: it has a complex zero at x = 2 - 5i, it has a real zero at x = -5, and it passes through the points specified in the table. This problem was a fantastic exercise in polynomial manipulation and problem-solving. We started with a seemingly simple set of information – a complex zero and a few table values – but we had to use a variety of techniques to piece together the puzzle. We leveraged the conjugate pair theorem to find the second complex zero, we used factoring to construct factors of the polynomial, and we set up a system of equations to solve for unknown coefficients and roots. But perhaps the most important lesson we learned is the value of perseverance. There were moments when we hit roadblocks, when our initial assumptions didn't pan out, and when we had to go back and re-evaluate our approach. But we didn't give up. We kept digging, kept exploring, and kept refining our solution until we found the answer. This is the essence of problem-solving in mathematics and in life. It's not always about getting the right answer on the first try; it's about the process of learning, adapting, and growing through the challenges we face. So, the next time you encounter a tough problem, remember this journey. Remember the importance of careful analysis, creative thinking, and unwavering persistence. And remember that even if you don't get it right away, the process of trying will make you a stronger, more resilient problem-solver. Well done, everyone! We conquered this polynomial beast together.