To determine which of the given options is a factor of the polynomial \( p(x) = x^3 + 2x^2 - 5x - 6 \), we can use the Factor Theorem, which states that if \( p(c) = 0 \) for some constant \( c \), then \( (x - c) \) is a factor of the polynomial.
Let's evaluate \( p(x) \) at the roots corresponding to the provided options:
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For \( x + 2 \) (i.e., \( c = -2 \)): \[ p(-2) = (-2)^3 + 2(-2)^2 - 5(-2) - 6 = -8 + 8 + 10 - 6 = 4 \quad (\text{not a factor}) \]
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For \( x - 3 \) (i.e., \( c = 3 \)): \[ p(3) = 3^3 + 2(3)^2 - 5(3) - 6 = 27 + 18 - 15 - 6 = 24 \quad (\text{not a factor}) \]
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For \( x + 1 \) (i.e., \( c = -1 \)): \[ p(-1) = (-1)^3 + 2(-1)^2 - 5(-1) - 6 = -1 + 2 + 5 - 6 = 0 \quad (\text{this is a factor}) \]
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For \( x - 1 \) (i.e., \( c = 1 \)): \[ p(1) = 1^3 + 2(1)^2 - 5(1) - 6 = 1 + 2 - 5 - 6 = -8 \quad (\text{not a factor}) \]
Therefore, the option that is a factor is:
\( x + 1 \)
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