Question

40. If x= 4/3 is a root of the polynomial
f(x) = 6x3 - 11x2 + kx - 20, find the value of k.
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Answer 1

Answer:

$f(x) = 6 {x}^{3} - 11 {x}^{2} + kx - 20 \\ x = \frac{4}{3} \\ 6 {( \frac{4}{3}) }^{3} - 11 { (\frac{4}{3}) }^{2} + k \times \frac{4}{3} - 20 = 0 \\ \frac{6 \times 64}{27} - \frac{11 \times 16}{9} + \frac{4k}{3} - 20 = 0 \\ \frac{384 - 176 \times 3 + 4k \times 9 - 20 \times 27}{27} = 0 \\ \frac{384 - 328 + 36k - 540}{27} = 0 \\ - 484 + 36k = 0 \\ 36k = 484 \\ k = 41$

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