Question

What is the remainder when x+x^9+x^25+x^49+x^81 is divided by x^3-x


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Answer 1

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p(x)=x+x^9+x^25+x^49+x^81

and divider d(x)= (x^3-x)=x(x^2–1)

Now p(x) / d(x) = x*( 1+x^8+x^24+x^48+x^80)/ x(x^2–1)

=( 1+x^8+x^24+x^48+x^80) / (x^2 – 1)

Thus we have to find out the remainder when 1+x^8+x^25+x^48+x^80 is divided by x^2–1

Let f(x)= 1+x^8+x^24+x^48+x^80……………..(1)

as the divider is of 2 degree thus remainder will be of degree 1

Let the remainder r(x)=ax+b………(2)

Then f(x)= (x^2–1)*q(x)+r(x)

OR r(x)= f(x)- (x–1)(x+1)*q(x) ………(3)

when f(x) is divided by x-1 then by remainder theorem from (3)

r(1)=f(1)

r(1)= 1+(1)^8 +(1)^24+(1)^48+(1)^80

r(1)=5

OR a*(1) +b=5 Or a+b=5…..(4)

when f(x) is divided by x+1 then from(3)

f(-1)= 0+r(-1) OR r(-1)=f(-1)

r(-1)= 1+(-1)^8 +(-1)^24+(-1)^48+(-1)^80

r(-1)=5

OR a*(-1) +b=5 Or -a+b=5……(5)

From(4) and (5) a=5 and b=0

Thus the remainder r(x)=5*x+0=5x

Thus the remainder is 5x

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