Math, asked by anureet5622, 3 months ago

Find the value of k for which the cubic polynomial 3y³-3/2y²+ky+5  is exactly divisible(y-1/2)​

Answers

Answered by karmaan958
22

Step-by-step explanation:

y - 1/2 = 0

y = 1/2

(putting the value of y = 1/2 in the expression

3y³-3/2y²+ky+5)

3y³-3/2y²+ky+5 = 0

3(1/2)³ - 3/2(1/2)² + k(1/2) + 5 = 0

3(1/8) - 3/2(1/4) + k(1/2) + 5 = 0

3/8 - 3/8 + k/2 + 5 = 0

k/2 + 5 = 0

k/2 = - 5

k = -5 × 2

k = -10

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