Question

If y = 4/3 is a zero of the polynomial p(y) =6y³ - 11y² + ky – 20, then the value of k is

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Answer 1

Answer:

Required Value of k is 19

Step-by-step explanation:

$p(y)=6y^{3}-11y^{2}+ky-20$

Since, y=4/3 is a zero of the polynomial p(y).

Therefore,

$p(\frac{4}{3} )=0\\=>6*(\frac{4}{3}) ^{3}-11*(\frac{4}{3}) ^{2}+k*(\frac{4}{3})-20=0\\=> \frac{384}{27} -\frac{176}{9}+\frac{4k}{3}-20=0\\=>\frac{384-528+36k-540}{27}=0\\=> -684+36k=0\\=>36k=684\\=>k=\frac{684}{36}\\=>k=19$

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