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if P(x)= ax^4 + bx^3 + cx^2 + dx + e has roots x= 1,2,3 and 4 and P(0) = 48, what is P(5)?
#48
#24
#0
#50
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- ans is 48....
x-1,x-2,x-3 ,x-4 are factors of p(x)
p(x) = k(x-1)(x-2)(x-3)(x-4) , where k is a constant
but given p(0) =48
24*k =48
k=2
p(x) =2(x-1)(x-2)(x-3)(x-4)
p(5)= 2*4*3*2*1 = 48 - 12 years agoHelpfull: Yes(27) No(0)
- answer a) 48
if 1 is a root then x-1 is a factor of p(x)
similarly x-2 is a factor,x-3 ,x-4 are factors
but p(x) is 4th degree polynomial therefore it can be in the form
p(x) = k(x-1)(x-2)(x-3)(x-4) , where k is a constant
but given p(0) =48
therefore 24 k =48
k=2
p(x) =2(x-1)(x-2)(x-3)(x-4)
p(5)= 2*4*3*2*1 = 48 - 12 years agoHelpfull: Yes(7) No(0)
- ans is 48
- 12 years agoHelpfull: Yes(0) No(0)
- ans. 48
- 12 years agoHelpfull: Yes(0) No(0)
- well think logically..........p(0)=48.....so as given epression p(x)=much much greater for x=1,2,3,4 pr 5.....look at the options....24,0 is nt possible as they r smaller than 48.....50is nt possible after looking at the expresion.....so only 48 is possible....we have to consider a=b=c=d=0 , only value of e=
- 12 years agoHelpfull: Yes(0) No(0)
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