To determine which of the given options is a factor of the polynomial \( f(x) = 2x^3 - 5x^2 - 68x + 35 \) using the Factor Theorem, we can substitute the values that correspond to the roots of each of the options into \( f(x) \). Recall that if \( (x - r) \) is a factor of \( f(x) \), then \( f(r) = 0 \).
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For \( (x + 5) \), we check \( f(-5) \): \[ f(-5) = 2(-5)^3 - 5(-5)^2 - 68(-5) + 35 = 2(-125) - 5(25) + 340 + 35 = -250 - 125 + 340 + 35 = 0 \] Thus, \( f(-5) = 0 \), meaning \( (x + 5) \) is a factor.
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For \( (x - 3) \), we check \( f(3) \): \[ f(3) = 2(3)^3 - 5(3)^2 - 68(3) + 35 = 2(27) - 5(9) - 204 + 35 = 54 - 45 - 204 + 35 = -160 \] Thus, \( f(3) \neq 0 \), so \( (x - 3) \) is not a factor.
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For \( (2x + 1) \), we check \( x = -\frac{1}{2} \): \[ f\left(-\frac{1}{2}\right) = 2\left(-\frac{1}{2}\right)^3 - 5\left(-\frac{1}{2}\right)^2 - 68\left(-\frac{1}{2}\right) + 35 \] \[ = 2(-\frac{1}{8}) - 5(\frac{1}{4}) + 34 + 35 \] \[ = -\frac{1}{4} - \frac{5}{4} + 34 + 35 = -\frac{6}{4} + 69 = -\frac{3}{2} + 69 \] This does not equal zero. Thus, \( (2x + 1) \) is not a factor.
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For \( (x + 7) \), we check \( f(-7) \): \[ f(-7) = 2(-7)^3 - 5(-7)^2 - 68(-7) + 35 \] \[ = 2(-343) - 5(49) + 476 + 35 = -686 - 245 + 476 + 35 \] \[ = -686 - 245 + 511 \] This does not equal zero. Thus, \( (x + 7) \) is not a factor.
Based on the above calculations, the factor of \( f(x) \) is:
\[ \boxed{(x + 5)} \]