The remainder when the polynomial 2 x^5-3 x^4 | vedclass

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(B) To find the remainder,we perform polynomial long division of $2 x^5-3 x^4+5 x^3-3 x^2+7 x-9$ by $x^2-x-3$:$1$. Divide $2 x^5$ by $x^2$ to get $2 x

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$41 x+3$

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(B) To find the remainder,we perform polynomial long division of $2 x^5-3 x^4+5 x^3-3 x^2+7 x-9$ by $x^2-x-3$:$1$. Divide $2 x^5$ by $x^2$ to get $2 x^3$. Multiply $(x^2-x-3)$ by $2 x^3$ to get $2 x^5-2 x^4-6 x^3$. Subtracting this from the dividend gives $-x^4+11 x^3-3 x^2+7 x-9$.$2$. Divide $-x^4$ by $x^2$ to get $-x^2$. Multiply $(x^2-x-3)$ by $-x^2$ to get $-x^4+x^3+3 x^2$. Subtracting this gives $10 x^3-6 x^2+7 x-9$.$3$. Divide $10 x^3$ by $x^2$ to get $10 x$. Multiply $(x^2-x-3)$ by $10 x$ to get $10 x^3-10 x^2-30 x$. Subtracting this gives $4 x^2+37 x-9$.$4$. Divide $4 x^2$ by $x^2$ to get $4$. Multiply $(x^2-x-3)$ by $4$ to get $4 x^2-4 x-12$. Subtracting this gives $41 x+3$.Thus,the remainder is $41 x+3$.

The remainder when the polynomial $2 x^5-3 x^4+5 x^3-3 x^2+7 x-9$ is divided by $x^2-x-3$ is

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