Math, asked by saishamahajan2, 3 months ago

1. Find the value of k for which the polynomial y 3 - 3y 2 + 3y + k has 3 as its zero.

Answers

Answered by tushargupta92
0

Let 3 is one of the zero of polynomial p(x) = -3y²+3y+k .

i.e. P(x) = -3y²-3y+k = 0

P(3) = (3)³-3(3)²+3(3)+k = 0

= 27-3(9)+9+k = 0

= 27-27+9+k = 0

= 9+k = 0

= k = -9

i.e. k = -9

Therefore, required value of k is -9.

Please select it brainliest answer.

Answered by madhav09062
0

Answer:

Let 3 is one of the zero of polynomial p(x) = y³-3y²+3y+k .

i.e. P(x) = y³-3y²-3y+k = 0

P(3) = (3)³-3(3)²+3(3)+k = 0

= 27-3(9)+9+k = 0

= 27-27+9+k = 0

= 9+k = 0

= k = -9

i.e. k = -9

Therefore, required value of k is -9.

Please select it brainliest answer.

Step-by-step explanation:

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