The factors of $2 x^{3} + 3 x^{2} - 3 x - 2$ is

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x + 2

x - 1

x + 1

x - \frac{1}{2}

answer is B, C.

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Detailed Solution

Given

2 x^{3} + 3 x^{2} - 3 x - 2

$f (1) = 2 {(1)}^{3} + 3 {(1)}^{2} - 3 (1) - 2$

$= 5 - 5 = 0$

$x - 1$ is factor of $2 x^{3} + 3 x^{2} - 3 x - 2$

$f (- 1) = 2 {(- 1)}^{3} + 3 {(- 1)}^{2} - 3 (- 1) - 2$

$= - 2 + 3 + 3 - 2$

$= - 4 + 6 = 2$

$x + 1$ is not a factor of $2 x^{3} + 3 x^{2} - 3 x - 2$

$f (- 2) = 2 {(- 2)}^{3} + 3 {(- 2)}^{2} - 3 (- 2) - 2$

$= - 16 + 12 + 6 - 2 = 18 - 18 = 0$

$x + 2$ is factor of $2 x^{3} + 3 x^{2} - 3 x - 2$

$f (\frac{1}{2}) = 2 {(\frac{1}{2})}^{3} + 3 {(\frac{1}{2})}^{2} - 3 (\frac{1}{2}) - 2$

$= 2. \frac{1}{8} + 3. \frac{1}{4} - \frac{3}{2} - 2 = \frac{1}{4} + \frac{3}{4} - \frac{3}{2} - 2 = - 1 - \frac{3}{2} = - \frac{5}{2}$

$x - \frac{1}{2}$ is not a factor of $2 x^{3} + 3 x^{2} - 3 x - 2$

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