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Asked by | 10 Mar, 2010, 10:18: AM

ATQ,

when p(x) is divided by x-2, it leaves a rem of 1

so we get

p(x)=(x-2)q(x)+1...(i)

next it's given that when p(x) is divided by x-3 rem is 3

i.e.

p(3)=3

substituting in (i) we get

p(3)=[1 x q(3)]+1

but p(3)=3

so

3=q(3)+1

so

q(3)=2

i.e. when q(x) is divided by x-3 it leaves a rem of 2

i.e.

q(x)=[(x-3)r(x)]+2... where r(x) is (say) the quotient

so substituting this value of q(x) in (i) we get,

p(x)=[(x-2)q(x)]+1

=([(x-2){(x-3)r(x)+2})+1

=[(x-2)(x-3)r(x)]+[2(x-2)]+1

so

p(x)=(x-2)(x-3)[r(x)]+[2x-4+1]

so the  reqd rem is 2x-4+1=2x-3

Answered by | 10 Mar, 2010, 04:35: PM
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