To determine which of the given polynomials is divisible by \(x - 2\) without a remainder, we can use polynomial long division for each polynomial. Alternatively, we can also use the Remainder Theorem, which states that if a polynomial \(f(x)\) is evaluated at \(x = c\), the remainder of the division of \(f(x)\) by \(x - c\) is simply \(f(c)\). If \(f(c) = 0\), then the polynomial is divisible by \(x - c\).
We will evaluate each polynomial at \(x = 2\):
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Polynomial: \(2x^3 - 5x^2 + 4x - 8\)
\[ f(2) = 2(2^3) - 5(2^2) + 4(2) - 8 \] \[ = 2(8) - 5(4) + 8 - 8 \] \[ = 16 - 20 + 8 - 8 = -4 \]
Not divisible by \(x - 2\).
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Polynomial: \(x^3 + 2x^2 - 5x - 6\)
\[ f(2) = (2^3) + 2(2^2) - 5(2) - 6 \] \[ = 8 + 2(4) - 10 - 6 \] \[ = 8 + 8 - 10 - 6 = 0 \]
Divisible by \(x - 2\).
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Polynomial: \(x^3 + 4x^2 - 6x + 8\)
\[ f(2) = (2^3) + 4(2^2) - 6(2) + 8 \] \[ = 8 + 4(4) - 12 + 8 \] \[ = 8 + 16 - 12 + 8 = 20 \]
Not divisible by \(x - 2\).
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Polynomial: \(3x^3 + 7x^2 - 2x + 5\)
\[ f(2) = 3(2^3) + 7(2^2) - 2(2) + 5 \] \[ = 3(8) + 7(4) - 4 + 5 \] \[ = 24 + 28 - 4 + 5 = 53 \]
Not divisible by \(x - 2\).
After evaluating all polynomials, we find that:
The polynomial \(x^3 + 2x^2 - 5x - 6\) is divisible by \(x - 2\) without a remainder.