Let f(x) =x^3+kx^2+hx+6. Find the value of h and k so that (x+1)and(x-2)are factor of f(x).
Answers
Answered by
29
Hii friend,
(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
(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
Answered by
1
Step-by-step explanation:
I hope you like this answer.
Attachments:

Similar questions
Physics,
9 months ago
Science,
9 months ago
Social Sciences,
1 year ago
India Languages,
1 year ago
English,
1 year ago