The factors of the polynomial $64 x^{3} - 56 x^{2} + 14 x - 1$ is

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$(8 x - 1)$

$x - 1$

$(4 x - 1)$

$(2 x - 1)$

answer is A, B.

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Detailed Solution

$64 x^{3} - 56 x^{2} + 14 x - 1$

$f (\frac{1}{2}) = 64 \times \frac{1}{8} - 56 \times \frac{1}{4} + 14 \times \frac{1}{2} - 1$

$= 8 - 14 + 7 - 1$

$= 15 - 15$

$= 0$ $\to 2 x - 1$ is a factor

$f (\frac{1}{4}) = 64 \times \frac{1}{64} - 56 \times \frac{1}{16} + 14 \times \frac{1}{4} - 1$

$= 1 - \frac{14}{4} + \frac{14}{4} - 1$

$= 0$ $\to 4 x - 1$ is a factor

$f (\frac{1}{8}) = 64 \times \frac{1}{64 \times 8} - 56 \times \frac{1}{64} + 14 \times \frac{1}{64} - 1$

$= \frac{1}{8} - \frac{7}{8} + \frac{7}{32} - 1$

$= \frac{- 6}{8} + \frac{7}{32} - 1$

$= \frac{- 24 + 7 - 32}{32}$ $\neq 0 \to$ $(8 x - 1)$ is not a factor

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