(1 - x) 2
Analyzing the Divisibility of the Algebraic Expression \(1-x-x^{n}+x^{n+1}\)
The question asks us to find which of the given options is a divisor of the expression \(1-x-x^{n}+x^{n+1}\), where \(n\) is a natural number. To solve this, we need to factor the given expression.
Step-by-Step Factorization of the Expression
Let the given expression be \(E\). We have:
\(E = 1 - x - x^n + x^{n+1}\)
We can group the terms to look for common factors:
\(E = (1 - x) + (-x^n + x^{n+1})\)
From the second group, we can factor out \(x^n\). Note that \(x^{n+1} = x^n \cdot x\). So, \(-x^n + x^{n+1} = -x^n(1 - x)\).
Substituting this back into the expression:
\(E = (1 - x) - x^n(1 - x)\)
Now, we can see that \((1 - x)\) is a common factor in both terms. We can factor out \((1 - x)\):
\(E = (1 - x)(1 - x^n)\)
Exploring Further Factorization using Algebraic Identity
We have factored the expression into \((1 - x)(1 - x^n)\). Now let's look at the term \((1 - x^n)\). We know a general algebraic identity for the difference of powers:
\(a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + \dots + ab^{n-2} + b^{n-1})\)
Using this identity for \(1 - x^n\) (where \(a=1\) and \(b=x\)), we get:
\(1^n - x^n = (1 - x)(1^{n-1} + 1^{n-2}x + \dots + 1x^{n-2} + x^{n-1})\)
Since \(1\) raised to any power is \(1\), this simplifies to:
\(1 - x^n = (1 - x)(1 + x + x^2 + \dots + x^{n-1})\)
This shows that for any natural number \(n\), \((1 - x^n)\) is divisible by \((1 - x)\).
Substituting back into the Original Expression
Now, substitute the factored form of \((1 - x^n)\) back into our factored expression \(E = (1 - x)(1 - x^n)\):
\(E = (1 - x) [(1 - x)(1 + x + x^2 + \dots + x^{n-1})]\)
\(E = (1 - x)^2 (1 + x + x^2 + \dots + x^{n-1})\)
Determining Divisibility based on the Factored Form
The expression \(1 - x - x^n + x^{n+1}\) has been factored into \((1 - x)^2\) multiplied by another factor \((1 + x + x^2 + \dots + x^{n-1})\). This means the original expression is a multiple of \((1 - x)^2\).
Therefore, the expression \(1 - x - x^n + x^{n+1}\) is divisible by \((1 - x)^2\).
Comparing with the Given Options
Let's check our result against the provided options:
- \((1 + x)^2\)
- \((1 - x)^2\)
- \(1 - 2x - x^2\)
- \(1 + 2x - x^2\)
Our factorization shows that the expression is divisible by \((1 - x)^2\). This matches option 2.
Conclusion
By factoring the expression \(1 - x - x^n + x^{n+1}\) as \((1 - x)^2 (1 + x + x^2 + \dots + x^{n-1})\), we see that it is always divisible by \((1 - x)^2\) for any natural number \(n\).
| Original Expression | Factored Form 1 | Factored Form 2 | Key Divisor Identified |
|---|---|---|---|
| \(1 - x - x^n + x^{n+1}\) | \((1 - x)(1 - x^n)\) | \((1 - x)^2 (1 + x + \dots + x^{n-1})\) | \((1 - x)^2\) |
Revision Table: Key Concepts for Divisibility and Factorization
| Concept | Description | Relevance to Problem |
|---|---|---|
| Factorization | Breaking down an expression into a product of simpler expressions. | Essential first step to identify divisors. |
| Divisibility | An expression A is divisible by expression B if A = B * C, where C is another expression (polynomial quotient). | The core concept tested in the problem. |
| Difference of Powers Identity | \(a^n - b^n = (a-b)(a^{n-1} + \dots + b^{n-1})\) | Used to factor \((1 - x^n)\) as \((1 - x)(1 + x + \dots + x^{n-1})\). |
| Natural Numbers (\(n\)) | The set of positive integers \(\{1, 2, 3, \dots\}\). | The identity \(1 - x^n\) being divisible by \((1-x)\) holds for all natural numbers \(n\). |
Additional Information: Understanding Factors and Divisors
In algebra, a divisor (or factor) of a polynomial \(P(x)\) is a polynomial \(D(x)\) such that \(P(x)\) can be written as \(P(x) = D(x) \cdot Q(x)\) for some polynomial \(Q(x)\). In this problem, we found that \(1 - x - x^n + x^{n+1}\) can be written as \((1 - x)^2 \cdot (1 + x + x^2 + \dots + x^{n-1})\). Here, \(P(x) = 1 - x - x^n + x^{n+1}\), \(D(x) = (1 - x)^2\), and \(Q(x) = 1 + x + x^2 + \dots + x^{n-1}\). This confirms that \((1 - x)^2\) is indeed a divisor.
The expression \(1 + x + x^2 + \dots + x^{n-1}\) is a geometric series sum. When \(n=1\), the original expression is \(1-x-x^1+x^2 = 1-2x+x^2 = (1-x)^2\). In this case, the divisor is \((1-x)^2\) and \(Q(x) = 1\). Our general formula \((1-x)^2 (1 + x + \dots + x^{n-1})\) gives \((1-x)^2 (1)\) for \(n=1\), which matches. When \(n=2\), the expression is \(1-x-x^2+x^3 = (1-x)(1-x^2) = (1-x)(1-x)(1+x) = (1-x)^2(1+x)\). Our formula gives \((1-x)^2 (1+x)\) for \(n=2\), which also matches. This confirms the factorization holds for different natural numbers \(n\).
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select the correct answer using the code given below:
