If polynomial ax^3+3x^2-3 and 2x^3 +5x-a leaves same remainder when each is divided by x-4,then find value of 'a'
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Answered by
4
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
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
revathikumarkan:
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Answered by
1
(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
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
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
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