To determine which of the given polynomials is divisible by \( x - 2 \) without a remainder, we can perform polynomial long division on each polynomial with \( x - 2 \) as the divisor, or we can evaluate each polynomial at \( x = 2 \). If substituting \( x = 2 \) into the polynomial results in zero, then the polynomial is divisible by \( x - 2 \).
We'll evaluate each polynomial at \( x = 2 \):
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For \( 2x^3 - 5x^2 + 4x - 8 \): \[ 2(2)^3 - 5(2)^2 + 4(2) - 8 = 2(8) - 5(4) + 8 - 8 = 16 - 20 + 8 - 8 = -4 \quad (\text{Not divisible}) \]
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For \( x^3 + 2x^2 - 5x - 6 \): \[ (2)^3 + 2(2)^2 - 5(2) - 6 = 8 + 2(4) - 10 - 6 = 8 + 8 - 10 - 6 = 0 \quad (\text{Divisible}) \]
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For \( x^3 + 4x^2 - 6x + 8 \): \[ (2)^3 + 4(2)^2 - 6(2) + 8 = 8 + 4(4) - 12 + 8 = 8 + 16 - 12 + 8 = 20 \quad (\text{Not divisible}) \]
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For \( 3x^3 + 7x^2 - 2x + 5 \): \[ 3(2)^3 + 7(2)^2 - 2(2) + 5 = 3(8) + 7(4) - 4 + 5 = 24 + 28 - 4 + 5 = 53 \quad (\text{Not divisible}) \]
Based on the evaluations, the only polynomial that is divisible by \( x - 2 \) without a remainder is:
\( x^3 + 2x^2 - 5x - 6 \).