Answers

2014-07-21T22:40:47+05:30
Put x=4 in both equation
and as the remainder is same so both will be equal 
so, 64a+48-3=128+20-a
      64a+a=148-45
      65a=103
       a=1.58
2 5 2
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2014-07-21T23:09:33+05:30
(ax^3+3x^2-3)/(x-4) = (2x^3+5x-a)/(x-4)
or, ax^3+3x-3=2x^3+5x-a
or, now when we put x=1(any constant),then
a+3-3=2+5-a
or, 2a=7
a = 7/2
0
it is giving other value for 2 . r u sure u hav done it correctly
oh,,sorry
we know that if two equation are equal then the ratio of x^2, x, and constant term are equal but when i follow it then i obtain two different values of a,finally i concluded an another constant should be b instead of a in eqn 2x^3+5X-a.now we have unique value of a=6/5
ratio of x^2 , x, and constant term not ratio of co efficient of x^2, x and constant term