We have #P(x)# an polynomial that meets the conditions:rest of dividing #P(x)# with #x-1# is 3 and #(x-1)P(x)+xP(x+2)=1#.How to find rest of dividing #P(x)# with #g=x^2-4x+3#?

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Answer 1 · 2017-06-12T09:56:23

#-x+4#

#P(x)=Q(x)g(x)+r(x) = Q(x)g(x) + ax+b#

and #g(x)=(x-1)(x-3)#

#P(1)=a+b= 3# and
#P(3)=3a+b#

but

#(1-1)P(1)+1P(3)=1 rArr P(3) = 1# then

#{(a+b=3),(3a+b=1):}#

Solving

#a=-1,b=4# and the division remainder is

#r(x)=-x+4#