Question
Given that b3+mb2 +nb + c is divisible by (b-s) if s3 + ms2 + ns + c = 0Also given that d3+d2+ed+1 is divisible by (d-1) and d3-4d2+fd-3 is divisible by (d-3)What is the value of e+f ?
Answer: Option C
:
C
Option (c)d3+d2+ed+1 is divisible by (d-1). As 1 is a root, Put d=1 in the equation and equate it to 0, to get the value of e1+1+e+1=0 e=-3Similarly put d=3 in d3-4d2+fd-3 to get the value of f 27-36+3f-3=0 3f=12 f=4e+f= -3+4=1
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:
C
Option (c)d3+d2+ed+1 is divisible by (d-1). As 1 is a root, Put d=1 in the equation and equate it to 0, to get the value of e1+1+e+1=0 e=-3Similarly put d=3 in d3-4d2+fd-3 to get the value of f 27-36+3f-3=0 3f=12 f=4e+f= -3+4=1
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