Question

If x = 4/3 is a zero of a polynomial p(x) = 6x^3 - 11x^2 + kx - 20, then find value of k.

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Answer 1

Answer:

$x = \frac{4}{3}$

$p(x) = {6x}^{3} - {11x}^{2} + kx - 20$

$p( \frac{4}{3} ) = 6( { \frac{4}{3} )}^{3} - 11( { \frac{4}{3}) }^{2} + k( \frac{4}{3} ) - 20 = 0$

$6( \frac{64}{27} ) - 11( \frac{16}{9} ) + \frac{4k}{3} - 20 = 0$

$\frac{384}{27} - \frac{176}{9} + \frac{4k}{3} - 20 = 0$

$\frac{(384 \times 1) +( - 176 \times 3) + (4k \times 9) + ( - 20 \times 27) }{27} = 0$

$\frac{384 - 528 + 36k - 540}{27} = 0$

$\frac{ - 684 + 36k}{27} = 0$

$- 684 + 36k = 0 \times 27$

$- 684 + 36k = 0$

$36k = 684$

$k = \frac{684}{36}$

$k = 19$

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Answer 2

Answer:

On putting the value of root of a polynomial into that polynomial we get 0

so, 6(4/3)^3-11(4/3)^2+k4/3-20=0

→128/9-176/9+k4/3 -20=0

→-48/9+k4/3 -20=0

→-16/3+k4/3 -20=0

→(4k-16)/3=20

→4k-16=60

→4k=76

→k=19