If $x^{2} + x + 1$ is a factor of the polynomial $3 x^{3} + 8 x^{2} + 8 x + 3 + 5 k$ , then the value of k is

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\frac{2}{5}

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\frac{5}{2}

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Solution

The correct option is B $\frac{2}{5}$
The long division & se remainder to zero
$x^{2} + x + 1) 3 x^{3} + 8 x^{2} + 8 x + 3 + 5 k (3 + 5 \frac{3 x^{3} + 3 x^{2} + 3 x}{\frac{5 {x/}^{2} + 5 x/ + 3}{5 {x/}^{2} + 5 x/ + 5}} - 2 + 5 k$
$∴$ Remainder =-2+5k=0
$\Rightarrow k = \frac{2}{5}$

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