if ax3+bx2+x-6 has x+2 as factor and leaves the remainder 4 when divided by (x-2), find the values of 'a' and 'b' .

it can be solved as follows :

 let p(X)=ax3+bx2+x-6

given that p(2)=4






==> a.-23+b.-22+(-2)-6=0



==>-8a+4b=8 (taking 4 common and dividing 8 by 4)

==>2a+b=2 ......(2)

on subtracting 1 from 2 we get,

  2a+b -(16a+b)=2-8

==> 2a+b-16a-b=-6


a=-6/-14 ==> a=3/7

hope the answer is correct..

on putting a in equation 1 we get,



  • -43

 sorry i also want the answer of the same question.If you get the answer please mail me.My g mail id is akkujalal@gmail.com

  • -17
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