Polynomial Solutions: Find P(x) With P(1)=8 & P(2)=68

Hey guys! Let's dive into the fascinating world of polynomials! Polynomials are fundamental in algebra, and understanding them is crucial for various mathematical applications. In this article, we're going to break down a specific type of polynomial problem, walking through each step to make sure you grasp the concepts. We'll focus on finding all polynomials that meet certain conditions, which involves using some cool algebraic techniques and logical reasoning. By the end of this guide, you'll be well-equipped to tackle similar problems and deepen your appreciation for the beauty of polynomials. So, let's get started and unlock the secrets of these mathematical expressions!

Let's get straight to the problem we're tackling. Imagine we have a polynomial $P(x)$ that looks something like this: $P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$. This might seem a bit intimidating, but don't worry, we'll break it down. Here, $n$ is a positive integer, and $a_n, a_{n-1}, \ldots, a_1, a_0$ are the coefficients of the polynomial, with $a_n$ not equal to zero. This last bit is important because it tells us that the degree of the polynomial is $n$. Now, this polynomial has to satisfy two special conditions:

1. The coefficients $a_0, a_1, \ldots, a_n$ are non-negative integers. This means we're only dealing with whole numbers that are zero or greater. No fractions or negative numbers allowed!
2. $P(1) = 8$ and $P(2) = 68$. These are the key pieces of information that will help us nail down the specific polynomial we're looking for. Basically, if we plug in $x = 1$ into the polynomial, we should get 8, and if we plug in $x = 2$, we should get 68. Think of these as clues that will guide us to the right answer.

Our mission, should we choose to accept it, is to find all such polynomials that meet these criteria. It might sound like a tough task, but we'll take it step by step, using some clever algebraic maneuvers and a bit of logical thinking. So, buckle up, and let's get ready to solve this polynomial puzzle!

Breaking Down the Conditions

Okay, guys, let's dig a little deeper into those conditions we talked about. The first condition, where the coefficients $a_0, a_1, \ldots, a_n$ are non-negative integers, is super important because it restricts the type of polynomials we can consider. If we didn't have this condition, the possibilities would be endless! But because we know we're only dealing with whole numbers that are zero or greater, it narrows down our search significantly. This means we won't be dealing with any fractions, decimals, or negative numbers in our coefficients, which is a huge help. It's like having a set of building blocks where each block is a whole number – we can only build our polynomial using these blocks.

The second condition, $P(1) = 8$ and $P(2) = 68$, gives us specific values that our polynomial must take at $x = 1$ and $x = 2$. These are like checkpoints that our polynomial needs to pass. When we substitute $x = 1$ into the polynomial equation, we get $a_n + a_{n-1} + \cdots + a_1 + a_0 = 8$. This tells us that the sum of all the coefficients must equal 8. This is a crucial piece of information because it puts a limit on how large the coefficients can be. Since they're all non-negative integers, none of them can be greater than 8. Now, when we substitute $x = 2$ into the polynomial equation, we get $a_n 2^n + a_{n-1} 2^{n-1} + \cdots + a_1 2 + a_0 = 68$. This is a more complex equation, but it provides another vital constraint on the coefficients. It tells us how the coefficients combine when weighted by powers of 2. Together, these two conditions give us a powerful system of equations that we can use to solve for the coefficients. It's like having two different perspectives on the same polynomial, and by combining these perspectives, we can zero in on the solutions. So, let's keep these conditions in mind as we move forward, because they're the key to unlocking this problem.

Alright, team, let's map out our strategy for tackling this polynomial problem. Our goal is to find all polynomials $P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$ that meet those two conditions we discussed: non-negative integer coefficients, and $P(1) = 8$ and $P(2) = 68$. So, how do we do it? First, let's leverage the condition $P(1) = 8$. This tells us that the sum of all the coefficients is 8, which is a fantastic starting point. We can write this as:

$$a_n + a_{n-1} + \cdots + a_1 + a_0 = 8$$

This equation gives us a good handle on the possible values of the coefficients. Since they are non-negative integers, we know that each $a_i$ must be between 0 and 8. Next, we'll use the condition $P(2) = 68$. This gives us another equation:

$$a_n 2^n + a_{n-1} 2^{n-1} + \cdots + a_1 2 + a_0 = 68$$

This equation is a bit trickier because it involves powers of 2. But it's also super useful because it tells us how the coefficients combine when weighted by powers of 2. The key here is to start thinking about how we can express 68 as a sum of powers of 2, keeping in mind that the coefficients must be non-negative integers and their sum must be 8. We can start by considering the largest power of 2 that is less than or equal to 68, which is $2^6 = 64$. This means that the highest possible degree of our polynomial is 6. However, we also need to remember that the sum of the coefficients must be 8. So, we'll need to carefully balance the powers of 2 to make sure we satisfy both conditions. Our approach will involve some clever trial and error, combined with a bit of logical deduction. We'll try different combinations of coefficients and powers of 2, checking to see if they satisfy both $P(1) = 8$ and $P(2) = 68$. It might sound like a bit of a puzzle, but that's what makes it fun! By systematically exploring the possibilities, we'll be able to find all the polynomials that fit the bill. So, let's roll up our sleeves and start digging into the details!

Finding Possible Polynomials

Okay, let's get our hands dirty and start hunting for those polynomials! We know that $P(1) = a_n + a_{n-1} + \cdots + a_1 + a_0 = 8$ and $P(2) = a_n 2^n + a_{n-1} 2^{n-1} + \cdots + a_1 2 + a_0 = 68$. Let's start by thinking about the highest possible degree of the polynomial. Since $2^6 = 64$ is less than 68, and $2^7 = 128$ is greater than 68, the highest degree we need to consider is 6. This means our polynomial could be of the form $P(x) = a_6 x^6 + a_5 x^5 + a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0$. But, we don't need to jump straight to degree 6. Let's start with simpler cases and work our way up.

Case 1: Degree 6

If we assume the degree is 6, then $a_6$ cannot be greater than 1, otherwise $P(2)$ will be greater than 68 since $2^6$ is already 64. If $a_6=1$, then we need to figure out how to get the remaining 4 (68-64) using coefficients $a_5$ through $a_0$. Also, the sum of the coefficients should be 8, and since we already have 1 from $a_6$, we have 7 left for the other coefficients. So, we need to find non-negative integers $a_5, a_4, a_3, a_2, a_1, a_0$ such that:

$$a_5 + a_4 + a_3 + a_2 + a_1 + a_0 = 7$$

$$a_5 2^5 + a_4 2^4 + a_3 2^3 + a_2 2^2 + a_1 2 + a_0 = 4$$

Now, let’s express the number 4 in binary format as that is 100 in base 2. So, 4 can be rewritten as $1 \cdot 2^2 + 0 \cdot 2^1 + 0 \cdot 2^0$. Comparing the equation $a_5 2^5 + a_4 2^4 + a_3 2^3 + a_2 2^2 + a_1 2 + a_0 = 4$, it indicates that we need one $2^2$, or that $a_2$ = 1, and the rest are 0. So, the coefficients $a_0$ to $a_5$ should sum to 7, and only $a_2$ is non-zero and equals 1. We have, $a_2 = 1$, so the other coefficients must sum to 6. Hence, $a_0 + a_1 + a_3 + a_4 + a_5 = 6$. All of them can be zero except one. So, let’s say $a_0 = 6$. So we have $P(x) = x^6 + x^2 + 6$. So this will work and we can proceed. This sounds like a good candidate.

Case 2: Lower Degrees

We can continue this process by considering lower degrees. If the degree is 5, the highest power of 2 is $2^5 = 32$. If the degree is 4, the highest power of 2 is $2^4 = 16$, and so on. We need to systematically explore these cases, keeping in mind the conditions $P(1) = 8$ and $P(2) = 68$. For each case, we'll try to find coefficients that satisfy both equations. This might involve a bit of trial and error, but that's part of the fun of solving these types of problems.

Verifying Solutions

Once we find a potential polynomial, we need to make sure it actually works. This means plugging in $x = 1$ and $x = 2$ and verifying that we get 8 and 68, respectively. It's like double-checking our work to make sure we haven't made any mistakes. This step is crucial because it ensures that our solutions are valid. If a polynomial doesn't satisfy both conditions, we need to discard it and keep searching. Verification is a key part of the problem-solving process, and it's important to be thorough and careful.

Okay, let's put our detective hats on and nail down some specific solutions. We've already started exploring the possibilities, and now it's time to get precise. Remember, we're looking for polynomials $P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$ where the coefficients are non-negative integers, and $P(1) = 8$ and $P(2) = 68$.

Solution 1

Let's revisit our degree 6 case, where we proposed $P(x) = x^6 + x^2 + 6$. Let's verify if this works. First, let’s compute P(1):

$$P(1) = 1^6 + 1^2 + 6 = 1 + 1 + 6 = 8$$

So, the condition $P(1) = 8$ is satisfied. Now, let's compute $P(2)$:

$$P(2) = 2^6 + 2^2 + 6 = 64 + 4 + 6 = 74$$

Oops! It looks like $P(2)$ is not equal to 68. So, our initial guess was close, but not quite right. This is a good reminder that verification is super important! We can't just assume a solution is correct; we need to check it. Let's backtrack and see where we might have gone wrong.

Solution 2

Let's try a different approach for a degree 6 polynomial. We still need $a_6 = 1$ to get close to 68. This means we have 4 remaining to account for in $P(2)$. Now, instead of putting that 4 into $x^2$, let's think about other combinations. What if we use the $x$ term? We need 4, so $2 a_1$ must be close to 4. So $a_1 = 2$. Now, we have accounted for the $x^6$ and the $2x$ terms, so 64+4 = 68. So we can proceed. We know that the sum of the coefficients should be 8. We already have 1 from $x^6$ and 2 from $2x$, so we need 5 more. So that could be $P(x) = x^6 + 2x + 5$, Let's test out.

$$P(1) = 1^6 + 2(1) + 5 = 1 + 2 + 5 = 8$$

$$P(2) = 2^6 + 2(2) + 5 = 64 + 4 + 5 = 73$$

Again, we failed the test, and $P(2)$ is not equal to 68. Let’s try setting $a_0=4$ to get $P(2)= 64+4=68$, and the coefficient should add up to 8, we have 1 from $a_6$ and 4 from $a_0$, so we need to add 3. If we set $a_1 = 3$, then $3(2)=6$. Then we can make $P(x)$ the following:

$$P(x) = x^6 + 3x + 4$$

Let’s test:

$$P(1) = 1^6 + 3(1) + 4 = 1 + 3 + 4 = 8$$

$$P(2) = 2^6 + 3(2) + 4 = 64 + 6 + 4 = 74$$

We are still not making it. It seems like going with degree 6 won’t work out. So let’s abandon this strategy and consider a new approach.

Solution 3

Let’s try a degree 5 polynomial: So, $P(x) = a_5 x^5 + a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0$, then we have $2^5 = 32$, and we need to reach 68. Let $a_5 = 2$, that gets us $2*32 = 64$. Now we are only 4 away from reaching 68. So, if we want to get a 4 from $x^2$ we can write $a_2 = 1$, and then the other coefficient is 0. Then we can make $P(x)$ the following:

$$P(x) = 2x^5 + x^2$$

We also need to verify $P(1) = 8$, where:

$$P(1) = 2(1^5) + 1^2 = 2 + 1 = 3$$

It doesn’t satisfy the $P(1)=8$. Also, the sum of the coefficients should be 8, but the current coefficient add up to 3. To make the sum equal to 8, we need to increase the coefficients by 5. Let $a_0 = 5$, so $P(1) = 2 + 1 + 5 = 8$, then $P(x)$ is defined as:

$$P(x) = 2x^5 + x^2 + 5$$

Let’s check $P(2)$ again:

$$P(2) = 2(2^5) + 2^2 + 5 = 2(32) + 4 + 5 = 64 + 4 + 5 = 73$$

Again, this doesn’t work either. Looks like using the $x^5$ is not a good idea. How about we focus on using the lower degrees.

Solution 4

Let’s try using $x^3$ and $x^2$. If we set $P(x) = a_3 x^3 + a_2 x^2 + a_1 x + a_0$, we want to satisfy: $a_3 2^3 + a_2 2^2 + a_1 2 + a_0 = 68$ and $a_3 + a_2 + a_1 + a_0 = 8$. We know that the $2^3$ equals 8, $2^2$ equals 4, so let’s figure out the numbers. 68 can be expressed as multiples of 8 plus a remainder. $68/8 = 8$ with a remainder of 4, but since we know the total number can be at most 8, it won’t work out. So let’s proceed with the next number. It means $8a_3 + 4a_2 + 2a_1 + a_0 = 68$ and $a_3 + a_2 + a_1 + a_0 = 8$. Try $a_3 = 8$, then we need $a_2, a_1$ and $a_0$ to be 0.

$$P(x) = 8x^3$$

Verify:

$$P(1) = 8(1^3) = 8$$

$$P(2) = 8(2^3) = 8(8) = 64$$

Since $P(2)$ is 64 instead of 68, this solution does not work.

Solution 5

Let’s try setting $a_3=7$. In this case, $8a_3$ is 56. $68 - 56$ is 12. Since $a_3 + a_2 + a_1 + a_0 = 8$, then $7 + a_2 + a_1 + a_0 = 8$. $a_2 + a_1 + a_0 = 1$. If $4a_2 + 2a_1 + a_0 = 12$, and $a_2 + a_1 + a_0 = 1$, and $12 > 4$, $12 > 2$, and $12 > 1$, so there is no way we can get this. Let’s consider this, what if we replace the $x^3$ with lower numbers.

How about 68 as a multiplication of $4$, then $68 / 4 = 17$. Let's try using $x^2$. It won’t work since the sum of the coefficient is 8, and we can reach 17. Let’s get the numbers again. $68 = 2^6 + 2^2 + 2^2$. So $68$ can be expressed as $2^6 + 4 + 4$ or $64 + 4 + 4$. We know that the $2^6$ won’t work. How about we use 17 with $2^2$ to make sure that number can reach 8. If we set $a_2=17$, then for $P(1) = 8$, it will exceed. Let’s say $a_1=34$, let’s say $2^1$. For $a_3$, let’s say 68/8 is smaller. If we get two numbers that get close to $68$, then we can get closer to the result. What about 2 different terms that satisfy.

After further exploration, we find another polynomial that fits the criteria:

$$P(x) = x^4 + x^3 + x^2 + 5$$

Let's check it:

$$P(1) = 1^4 + 1^3 + 1^2 + 5 = 1 + 1 + 1 + 5 = 8$$

$$P(2) = 2^4 + 2^3 + 2^2 + 5 = 16 + 8 + 4 + 5 = 33$$

It doesn’t work either.

Solution 6

If we analyze the equation again, $a_n 2^n + a_{n-1} 2^{n-1} + \cdots + a_1 2 + a_0 = 68$ and $a_n + a_{n-1} + \cdots + a_1 + a_0 = 8$. Let’s try $x^5$, we know the maximum is 32, but if we get $2(32)$, we get 64. So $P(x) = 2x^5$, but to get it to 68, we need 4. 4 is less than 8, so we can try the last term to solve that number. Then we can use $x^0$, so $P(x) = a_5x^5 + a_0 = 68$, $2(2^5) + a_0 = 68$, then $a_0 = 4$. So, $P(x) = 2x^5 + 4$. $P(1) = 2(1)^5 + 4 = 6$, it won’t work either. We need to hit 8, and the sum coefficient is 6. What if we add 2 to $x$, then $P(x)=2x^5 + 2x + 4$.

Let's check it:

$$P(1) = 2 + 2 + 4 = 8$$

$$P(2) = 2(32) + 2(2) + 4 = 64 + 4 + 4 = 72$$

This is way too high.

Solution 7

Let’s try $x^4$. With $x^4$ we get 16, but to reach 68, we need to find another number to add up. if we divide $68/16$, it goes to 4 with a remainder. Let’s try $4 x^4$, which gives 64. We need to add $4$, or $4(1)$, so $P(x) = 4x^4 + 4$. $P(1) = 4 + 4 = 8$. Check. $P(2) = 4(16) + 4 = 64 + 4 = 68$!

So, we have a winner!

$$P(x) = 4x^4 + 4$$

Well, guys, we did it! We've navigated the world of polynomials, tackled a challenging problem, and emerged victorious. It was a journey filled with algebraic manipulations, logical deductions, and a fair bit of trial and error. But in the end, we found a polynomial that satisfies all the given conditions: $P(x) = 4x^4 + 4$. This problem highlights the beauty and power of polynomials in mathematics. By understanding their properties and using clever techniques, we can solve complex equations and uncover hidden relationships. So, keep exploring, keep questioning, and keep having fun with math! Who knows what other mathematical treasures you'll discover?