Answer :
Answer:
To determine which of the given polynomials has \( (x+1) \) as a factor, we can use the Remainder Theorem. According to the Remainder Theorem, if \( (x+1) \) is a factor of a polynomial \( P(x) \), then \( P(-1) = 0 \).
Let's evaluate each polynomial at \( x = -1 \) to check which one has \( (x+1) \) as a factor:
1. \( P(x) = x^3 + x^2 + x + 1 \)
\[
P(-1) = (-1)^3 + (-1)^2 + (-1) + 1 = -1 + 1 - 1 + 1 = 0
\]
Since \( P(-1) = 0 \), \( (x+1) \) is a factor of \( x^3 + x^2 + x + 1 \).
2. \( P(x) = x^4 + x^3 + x^2 + x + 1 \)
\[
P(-1) = (-1)^4 + (-1)^3 + (-1)^2 + (-1) + 1 = 1 - 1 + 1 - 1 + 1 = 1
\]
\( P(-1) \neq 0 \), so \( (x+1) \) is not a factor of \( x^4 + x^3 + x^2 + x + 1 \).
3. \( P(x) = x^4 + 3x^3 + 3x^2 + x + 1 \)
\[
P(-1) = (-1)^4 + 3(-1)^3 + 3(-1)^2 + (-1) + 1 = 1 - 3 + 3 - 1 + 1 = 1
\]
\( P(-1) \neq 0 \), so \( (x+1) \) is not a factor of \( x^4 + 3x^3 + 3x^2 + x + 1 \).
4. \( P(x) = x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2} \)
\[
P(-1) = (-1)^3 - (-1)^2 - (2 + \sqrt{2})(-1) + \sqrt{2} = -1 - 1 + 2 + \sqrt{2} + \sqrt{2} = -2 + 2\sqrt{2}
\]
\( P(-1) \neq 0 \), so \( (x+1) \) is not a factor of \( x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2} \).
Therefore, the polynomial that has \( (x+1) \) as a factor is \( \boxed{x^3 + x^2 + x + 1} \).
Step-by-step explanation:
(a) option
Sol: x+1=0
x= -1
p(-1)= (-1)³ + (-1)² + (-1) + 1
= -1 + 1 -1 +1
= 0
Since , the remainder is 0 so so x³+x²+x+1 is a factor of (x+1). i.e (a) part