If (x-2) is a zero of x3-4x2-kx-8,find k
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Answered by
19
Let p (x) = x^3 - 4x^2 - kx - 8
If x - 2 is a factor of p (x) , then p (2) = 0
=> 2^3 - 4 × 2^2 - k × 2 - 8 = 0
=> 8 - 16 -2k - 8 = 0
=> -2k - 16 = 0
=> -2k = 16
=> k = 16/-2
=> k = -8
Hope it helps.
If x - 2 is a factor of p (x) , then p (2) = 0
=> 2^3 - 4 × 2^2 - k × 2 - 8 = 0
=> 8 - 16 -2k - 8 = 0
=> -2k - 16 = 0
=> -2k = 16
=> k = 16/-2
=> k = -8
Hope it helps.
Answered by
6
x-2=0
x=2
p(x) =x^3-4x^2-kx-8
p(2) =(2) ^3-4*(2) ^2-k*2-8=0
8-4*4-2k-8=0
8-16-2k-8=0
-8-2k-8=0
-16-2k=0
-2k=16
k=-16/2
k=-8
hope it will help. plz mark it as brainlist plz
x=2
p(x) =x^3-4x^2-kx-8
p(2) =(2) ^3-4*(2) ^2-k*2-8=0
8-4*4-2k-8=0
8-16-2k-8=0
-8-2k-8=0
-16-2k=0
-2k=16
k=-16/2
k=-8
hope it will help. plz mark it as brainlist plz
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