If $x = \frac{4}{3}$ is a root of the polynomial
$f (x) = 6 x^{3} - 11 x^{2} + k x - 20$ , then find the value of $k$ .

15

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19

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22

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26

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Solution

The correct option is B $19$
$f (x) = 6 x^{3} - 11 x^{2} + k x - 20$
$f (\frac{4}{3}) = 6 {(\frac{4}{3})}^{3} - 11 {(\frac{4}{3})}^{2} + k (\frac{4}{3}) - 20 = 0$
$\Rightarrow 6 \cdot \frac{64}{27} - 11 \cdot \frac{16}{9} + \frac{4 k}{3} - 20 = 0$ ......... $(multiplying both side by 9 , we get)$
$\Rightarrow 128 - 176 + 12 k - 180 = 0$
$\Rightarrow 12 k + 128 - 356 = 0 \Rightarrow 12 k = 128 \Rightarrow k = 19$

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