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

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Solution

When we put the value of root of a polynomial into that polynomial we get $0$
so,
$6 (\frac{4}{3})^{3} - 11 (\frac{4}{3})^{2} + k \frac{4}{3} - 20 = 0$

$= \frac{128}{9} - \frac{176}{9} + k \frac{4}{3} - 20 = 0$

$= \frac{- 48}{9} + k \frac{4}{3} - 20 = 0$

$\frac{- 48}{9} + \frac{4}{3} k - 20 = 0$

$= \frac{- 48 + 12 k - 180}{9} = 0$

$= - 48 - 180 + 12 k = 0$

$= - 228 + 12 k$

$12 k = 228$

$k = \frac{228}{12}$

$k = 19$

This is the required solution.

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