Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Ask your questions and receive accurate answers from professionals with extensive experience in various fields on our platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
To determine which statement is correct when [tex]\(x + 1\)[/tex] is a factor of the polynomial [tex]\(f(x) = a_0 x^n + a_1 x^{n-1} + \ldots + a_n\)[/tex], we start by noting a fundamental property of polynomials:
If [tex]\(x + 1\)[/tex] is a factor of [tex]\(f(x)\)[/tex], then the polynomial [tex]\(f(x)\)[/tex] evaluated at [tex]\(x = -1\)[/tex] must be zero. This is derived from the factor theorem which states that if [tex]\((x - c)\)[/tex] is a factor of a polynomial [tex]\(f(x)\)[/tex], then [tex]\(f(c) = 0\)[/tex].
Given:
[tex]\[ f(x) = a_0 x^n + a_1 x^{n-1} + a_2 x^{n-2} + \ldots + a_{n-1} x + a_n \][/tex]
We evaluate [tex]\( f(x) \)[/tex] at [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = a_0 (-1)^n + a_1 (-1)^{n-1} + a_2 (-1)^{n-2} + \ldots + a_{n-1} (-1) + a_n \][/tex]
Since [tex]\( x + 1 \)[/tex] is a factor, [tex]\( f(-1) \)[/tex] must equal zero:
[tex]\[ a_0 (-1)^n + a_1 (-1)^{n-1} + a_2 (-1)^{n-2} + \ldots + a_{n-1} (-1) + a_n = 0 \][/tex]
This implies a sum of the coefficients, considering the alternating signs of the terms. Notice that the signs alternate based on the power of [tex]\(-1\)[/tex]:
For even [tex]\( n \)[/tex]:
[tex]\[ a_0 - a_1 + a_2 - a_3 + \ldots + a_{n-1} - a_n = 0 \][/tex]
For odd [tex]\( n \)[/tex]:
[tex]\[ a_0 - a_1 + a_2 - a_3 + \ldots - a_{n-1} + a_n = 0 \][/tex]
This equation consolidates the contributions of all coefficients and dictates the relationship we need to examine.
Now let's evaluate each given statement:
1. [tex]\( a_1 + a_3 + a_5 + \ldots = a_0 + a_2 + a_4 + \ldots \)[/tex]
- This does not directly follow from our alternating sum equation.
2. [tex]\( a_0 + a_1 + a_2 + a_3 + \ldots = 0 \)[/tex]
- This is indeed true based on our equation, as it implies the sum of all coefficients (with the same signs) results in zero.
3. [tex]\( a^2 + b + 1 = 0 \)[/tex]
- This has no relevant connection to the polynomial relationship described.
4. [tex]\( a_1 - a_2 - a_3 - \ldots = 0 \)[/tex]
- This would need to include alternating signs correctly to match our earlier equation but doesn't.
Therefore, the correct statement is:
[tex]\[ \boxed{2} \][/tex]
If [tex]\(x + 1\)[/tex] is a factor of [tex]\(f(x)\)[/tex], then the polynomial [tex]\(f(x)\)[/tex] evaluated at [tex]\(x = -1\)[/tex] must be zero. This is derived from the factor theorem which states that if [tex]\((x - c)\)[/tex] is a factor of a polynomial [tex]\(f(x)\)[/tex], then [tex]\(f(c) = 0\)[/tex].
Given:
[tex]\[ f(x) = a_0 x^n + a_1 x^{n-1} + a_2 x^{n-2} + \ldots + a_{n-1} x + a_n \][/tex]
We evaluate [tex]\( f(x) \)[/tex] at [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = a_0 (-1)^n + a_1 (-1)^{n-1} + a_2 (-1)^{n-2} + \ldots + a_{n-1} (-1) + a_n \][/tex]
Since [tex]\( x + 1 \)[/tex] is a factor, [tex]\( f(-1) \)[/tex] must equal zero:
[tex]\[ a_0 (-1)^n + a_1 (-1)^{n-1} + a_2 (-1)^{n-2} + \ldots + a_{n-1} (-1) + a_n = 0 \][/tex]
This implies a sum of the coefficients, considering the alternating signs of the terms. Notice that the signs alternate based on the power of [tex]\(-1\)[/tex]:
For even [tex]\( n \)[/tex]:
[tex]\[ a_0 - a_1 + a_2 - a_3 + \ldots + a_{n-1} - a_n = 0 \][/tex]
For odd [tex]\( n \)[/tex]:
[tex]\[ a_0 - a_1 + a_2 - a_3 + \ldots - a_{n-1} + a_n = 0 \][/tex]
This equation consolidates the contributions of all coefficients and dictates the relationship we need to examine.
Now let's evaluate each given statement:
1. [tex]\( a_1 + a_3 + a_5 + \ldots = a_0 + a_2 + a_4 + \ldots \)[/tex]
- This does not directly follow from our alternating sum equation.
2. [tex]\( a_0 + a_1 + a_2 + a_3 + \ldots = 0 \)[/tex]
- This is indeed true based on our equation, as it implies the sum of all coefficients (with the same signs) results in zero.
3. [tex]\( a^2 + b + 1 = 0 \)[/tex]
- This has no relevant connection to the polynomial relationship described.
4. [tex]\( a_1 - a_2 - a_3 - \ldots = 0 \)[/tex]
- This would need to include alternating signs correctly to match our earlier equation but doesn't.
Therefore, the correct statement is:
[tex]\[ \boxed{2} \][/tex]
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.
