To determine which of the given options is a factor of \( f(x) = (x+5)(2x^2 - 15x + 7) \) using the Factor Theorem, we need to evaluate \( f(x) \) at the possible roots of the polynomial formed by substituting \( x \) values into the factors.
The candidates that can be roots of the factors are:
- \( x - 7 \) (root at \( x = 7 \))
- \( x - 3 \) (root at \( x = 3 \))
- \( x + 7 \) (root at \( x = -7 \))
- \( 2x + 1 \) (root at \( x = -\frac{1}{2} \))
We will evaluate \( f(x) \) at these points.
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For \( x = 7 \): \[ f(7) = (7 + 5)(2(7)^2 - 15(7) + 7) = 12(2(49) - 105 + 7) = 12(98 - 105 + 7) = 12(0) = 0 \] So, \( x - 7 \) is a factor.
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For \( x = 3 \): \[ f(3) = (3 + 5)(2(3)^2 - 15(3) + 7) = 8(2(9) - 45 + 7) = 8(18 - 45 + 7) = 8(-20) = -160 \neq 0 \]
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For \( x = -7 \): \[ f(-7) = (-7 + 5)(2(-7)^2 - 15(-7) + 7) = (-2)(2(49) + 105 + 7) = (-2)(98 + 105 + 7) = -2(210) = -420 \neq 0 \]
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For \( x = -\frac{1}{2} \): \[ f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2} + 5\right)\left(2\left(-\frac{1}{2}\right)^2 - 15\left(-\frac{1}{2}\right) + 7\right) = \left(\frac{9}{2}\right)\left(2\left(\frac{1}{4}\right) + \frac{15}{2} + 7\right) = \left(\frac{9}{2}\right)\left(\frac{1}{2} + \frac{15}{2} + \frac{14}{2}\right) = \left(\frac{9}{2}\right)\left(\frac{30}{2}\right) = \frac{9 \cdot 30}{4} \neq 0 \]
From this evaluation, we find that \( f(7) = 0 \), meaning that \( (x - 7) \) is the only factor among the options listed.
Thus, the factor of \( f(x) \) is: \[ \boxed{(x - 7)} \]
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