plzz solve this ques
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the value of k is -3
hope this helps you..
hope this helps you..
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Given Dividend = 2x^3 - kx^2 + 3x + 10.
Given Divisior = x + 2.
Put divisor = 0
= > x + 2 = 0
= > x = -2.
Let f(x) = 2x^3 - kx^2 + 3x + 10
Substitute x = -2 in f(x), we get :
f(-2) = 2(-2)^3 - k(-2)^2 + 3(-2) + 10 = 0
= -16 - 4k - 6 + 10 = 0
= -4k - 12
= k = -3.
Therefore the value of k = -3
Hope this helps!
Given Divisior = x + 2.
Put divisor = 0
= > x + 2 = 0
= > x = -2.
Let f(x) = 2x^3 - kx^2 + 3x + 10
Substitute x = -2 in f(x), we get :
f(-2) = 2(-2)^3 - k(-2)^2 + 3(-2) + 10 = 0
= -16 - 4k - 6 + 10 = 0
= -4k - 12
= k = -3.
Therefore the value of k = -3
Hope this helps!
Arooza:
sorryy ...this is wrong ans
Edited..Thanks
okk
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