Question

let f(x)=x³+kx²+hx+6. find the value of h and k so that (x+1) and (x-2) are factors of f(x)

Answer 1

(X+1) and (X-2) are the factor of the polynomial X³+KX²+HX+6.

(X+1) = 0 OR (X-2) = 0

=> X = -1 OR X= 2

F(X) = X³+KX²+HX+6

F(-1) = (-1)³ + K × (-1)² + H × -1 + 6

=> -1 +K -H +6 = 0

=> K - H +5 = 0

=>K - H = -5.......(1)

Again,

F(X) = X³+KX²+HX+6

F(2) = (2)³ +K × (2)² + H × 2 + 6

=> 8 +4K +2H +6 = 0

=> 4K + 2H +14 = 0

=> 4K +2H = -14.....(2)

From equation (1) we get,

K - H = -5

K = -5 + H........(3)

Putting the value of K in equation (2)

=> 4K + 2H = -14

=> 4 × (-5+H) +2H = -14

=> -20 + 4H +2H = -14

=> 6H = -14+20

=> H = 6/6 = 1

Putting the value of H in equation (3)

K = -5 + H

K = -5 +1 = 4.

Hence,

H = 1 and K = 4

HOPE IT WILL HELP YOU

Answer 2

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Explanation:

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