Let r(x) be the remainder when the polynomial | vedclass

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(C) Let $p(x) = x^{135}+x^{125}-x^{115}+x^5+1$ and $q(x) = x^3-x = x(x-1)(x+1)$.Since the divisor $q(x)$ is of degree $3$,the remainder $r(x)$ will be

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degree of $r(x)$ is one

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(C) Let $p(x) = x^{135}+x^{125}-x^{115}+x^5+1$ and $q(x) = x^3-x = x(x-1)(x+1)$.Since the divisor $q(x)$ is of degree $3$,the remainder $r(x)$ will be of the form $ax^2+bx+c$.Thus,$p(x) = (x^3-x)k(x) + ax^2+bx+c$.For $x=0$: $p(0) = 0+0-0+0+1 = 1$. So,$c = 1$.For $x=1$: $p(1) = 1+1-1+1+1 = 3$. So,$a+b+c = 3 \Rightarrow a+b = 2$.For $x=-1$: $p(-1) = (-1)^{135}+(-1)^{125}-(-1)^{115}+(-1)^5+1 = -1-1+1-1+1 = -1$. So,$a-b+c = -1 \Rightarrow a-b = -2$.Adding the two equations: $2a = 0 \Rightarrow a = 0$.Subtracting the two equations: $2b = 4 \Rightarrow b = 2$.Therefore,$r(x) = 0x^2 + 2x + 1 = 2x+1$.The degree of $r(x) = 2x+1$ is $1$.

Let $r(x)$ be the remainder when the polynomial $x^{135}+x^{125}-x^{115}+x^5+1$ is divided by $x^3-x$. Then,

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