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Let [tex]$f(x)=a_0 x^n+a_1 x^{n-1}+\ldots+a_n$[/tex] (where [tex]$a_0 \neq 0$[/tex]) be a polynomial of degree [tex]$n$[/tex]. If [tex]$x+1$[/tex] is one of its factors, then:

A. [tex]$a_1 + a_3 + a_5 + \ldots = a_0 + a_2 + a_4 + \ldots$[/tex]
B. [tex]$a_0 + a_1 + a_2 + a_3 + \ldots = 0$[/tex]
C. [tex]$a^2 + b + 1 = 0$[/tex]
D. [tex]$a_1 - a_2 - a_3 - \ldots = 0$[/tex]

Sagot :

GinnyAnsweridn
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]
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