Consider the following statements in respect of the polynomial 1 - x - xn + xn+1 where n is a natural number : 1. It is divisible by 1 - 2x + x2. 2. It is divisible by 1 - xn. Which of the statements given above is/are correct ?
Both 1 and 2
Analyzing Polynomial Divisibility: \(1 - x - x^n + x^{n+1}\)
We are given the polynomial \(P(x) = 1 - x - x^n + x^{n+1}\), where \(n\) is a natural number. We need to examine two statements regarding its divisibility.
Statement 1: Divisibility by \(1 - 2x + x^2\)
The first statement says that the polynomial \(P(x)\) is divisible by \(1 - 2x + x^2\). Let's factor the divisor:
\(1 - 2x + x^2 = (1 - x)^2\)
For a polynomial to be divisible by \((1-x)^2\), it must have \(x=1\) as a root with a multiplicity of at least 2. This means both the polynomial itself and its first derivative must be zero at \(x=1\).
Let's evaluate \(P(x)\) at \(x=1\):
\(P(1) = 1 - 1 - 1^n + 1^{n+1}\)
Since \(n\) is a natural number (\(n \ge 1\)), \(1^n = 1\) and \(1^{n+1} = 1\).
\(P(1) = 1 - 1 - 1 + 1 = 0\)
So, \(x=1\) is a root of \(P(x)\), meaning \(P(x)\) is divisible by \((1-x)\).
Now, let's find the first derivative of \(P(x)\) and evaluate it at \(x=1\).
\(P(x) = 1 - x - x^n + x^{n+1}\)
\(P'(x) = \frac{d}{dx}(1) - \frac{d}{dx}(x) - \frac{d}{dx}(x^n) + \frac{d}{dx}(x^{n+1})\)
\(P'(x) = 0 - 1 - nx^{n-1} + (n+1)x^n\)
Now, evaluate \(P'(1)\):
\(P'(1) = -1 - n(1)^{n-1} + (n+1)(1)^n\)
Since \(n\) is a natural number, \(1^{n-1} = 1\) and \(1^n = 1\).
\(P'(1) = -1 - n(1) + (n+1)(1)\)
\(P'(1) = -1 - n + n + 1 = 0\)
Since \(P(1) = 0\) and \(P'(1) = 0\), \(x=1\) is a root of multiplicity at least 2. This implies that \(P(x)\) is divisible by \((x-1)^2\). Note that \((x-1)^2 = (- (1-x))^2 = (1-x)^2\). Therefore, \(P(x)\) is divisible by \(1 - 2x + x^2\).
Statement 1 is correct.
Statement 2: Divisibility by \(1 - x^n\)
The second statement says that the polynomial \(P(x)\) is divisible by \(1 - x^n\). Let's try to factor the polynomial \(P(x)\) directly.
\(P(x) = 1 - x - x^n + x^{n+1}\)
We can group terms to find common factors:
\(P(x) = (1 - x) + (- x^n + x^{n+1})\)
Factor out \(-x^n\) from the second group:
\(P(x) = (1 - x) - x^n(1 - x)\)
Now, we see that \((1-x)\) is a common factor in both terms:
\(P(x) = (1 - x)(1 - x^n)\)
Since \(P(x)\) can be written as the product of \((1-x)\) and \((1-x^n)\), it is clearly divisible by \((1 - x^n)\).
Statement 2 is correct.
Conclusion on Polynomial Divisibility Statements
Based on our analysis, both Statement 1 (divisibility by \(1 - 2x + x^2\)) and Statement 2 (divisibility by \(1 - x^n\)) are correct for the polynomial \(1 - x - x^n + x^{n+1}\) where \(n\) is a natural number.
| Statement | Divisor | Verification Method | Result |
|---|---|---|---|
| 1 | \(1 - 2x + x^2 = (1-x)^2\) | Check if \(P(1)=0\) and \(P'(1)=0\). | \(P(1)=0\), \(P'(1)=0\). Divisible. |
| 2 | \(1 - x^n\) | Factor the polynomial \(P(x)\). | \(P(x) = (1-x)(1-x^n)\). Divisible. |
Revision Table: Key Concepts for Polynomial Divisibility
| Concept | Description | Relevance to the Problem |
|---|---|---|
| Polynomial Divisibility | A polynomial \(A(x)\) is divisible by \(B(x)\) if \(A(x) = B(x) \cdot Q(x)\) for some polynomial \(Q(x)\). | We checked if \(P(x)\) could be expressed as the product of the given divisors and another polynomial. |
| Factor Theorem | If \(P(a) = 0\), then \((x-a)\) is a factor of \(P(x)\). | Used in Statement 1: \(P(1)=0\) implies \((x-1)\) is a factor. |
| Root Multiplicity | A root \(a\) has multiplicity \(m\) if \((x-a)^m\) is a factor of \(P(x)\) but \((x-a)^{m+1}\) is not. If \(P(a)=0, P'(a)=0, ..., P^{(m-1)}(a)=0\) but \(P^{(m)}(a) \ne 0\), the root \(a\) has multiplicity \(m\). | Used in Statement 1: \(P(1)=0\) and \(P'(1)=0\) implies \(x=1\) is a root of multiplicity at least 2, so \((x-1)^2\) is a factor. |
| Polynomial Factorization | Breaking down a polynomial into a product of simpler polynomials (factors). | Used in Statement 2 by factoring \(P(x)\) to explicitly show \((1-x^n)\) as a factor. |
Additional Information: Polynomial Properties and Divisibility
Understanding polynomial properties is crucial for solving divisibility problems. Key ideas include the Factor Theorem, the Remainder Theorem (which states that the remainder of the division of a polynomial \(P(x)\) by \((x-a)\) is \(P(a)\)), and the concept of root multiplicity.
When a polynomial is divisible by a squared factor like \((x-a)^2\), it signifies that the root \(a\) is repeated. This can be checked by evaluating the polynomial and its derivatives at the root \(a\).
Algebraic manipulation, such as grouping terms and factoring, is a powerful technique to reveal the structure of a polynomial and identify its factors, as demonstrated in the analysis of Statement 2.
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