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Problem Of The Week 51

with 4 comments


Let P(x)=ax^4 +bx^3+cx^2+dx+e be a polynomial with all integer coefficients and a=1. If√2+√5 is one of the roots of P(x)=0 , which of the following can be the value of |(b+c+d+e)|?

1) 103

2) 89

3) 63

4) 23

5) 5

Written by Implex

October 9, 2008 at 7:24 pm

4 Responses

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  1. is the answer d) 23
    |(b+c+d+e)|=23

    b=d=0, c= -18 e=41

    correct me??

    Ankaj

    October 11, 2008 at 5:57 pm

  2. aakaj its 5. yaar

    Varun

    October 11, 2008 at 11:05 pm

  3. hi Varun,

    correct me…this is how we are solving…put the value of x in the expression and then equate p(x) to zero….compare the terms with zero….

    the values which i get are b=0, d=0, e=41 and c=-18…. (correct me if these values are incorrect)…

    did a recalculation…not sure where i went wrong😦

    ankaj

    October 12, 2008 at 2:50 am

  4. 4.The values are b=0,c=-14,d=0 and e=9..The equations are
    28+2c=0
    11b+d=0
    17b+d=0
    89+e+7c=0
    Answer choice e)

    Celebrating Life

    October 13, 2008 at 8:52 am


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