What must be added to the polynomial p(x) = 6 | vedclass

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(B) To find what must be added to $p(x)$ to make it divisible by $g(x) = 3x^2 - 2x + 4$,we perform polynomial long division of $p(x)$ by $g(x)$.Dividi

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$17x - 13$

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(B) To find what must be added to $p(x)$ to make it divisible by $g(x) = 3x^2 - 2x + 4$,we perform polynomial long division of $p(x)$ by $g(x)$.Dividing $6x^5 + 5x^4 + 11x^3 - 3x^2 + x + 1$ by $3x^2 - 2x + 4$:$1$. Divide $6x^5$ by $3x^2$ to get $2x^3$. Multiply $2x^3(3x^2 - 2x + 4) = 6x^5 - 4x^4 + 8x^3$. Subtracting this from $p(x)$ gives $9x^4 + 3x^3 - 3x^2 + x + 1$.$2$. Divide $9x^4$ by $3x^2$ to get $3x^2$. Multiply $3x^2(3x^2 - 2x + 4) = 9x^4 - 6x^3 + 12x^2$. Subtracting this gives $9x^3 - 15x^2 + x + 1$.$3$. Divide $9x^3$ by $3x^2$ to get $3x$. Multiply $3x(3x^2 - 2x + 4) = 9x^3 - 6x^2 + 12x$. Subtracting this gives $-9x^2 - 11x + 1$.$4$. Divide $-9x^2$ by $3x^2$ to get $-3$. Multiply $-3(3x^2 - 2x + 4) = -9x^2 + 6x - 12$. Subtracting this gives $-17x + 13$.The remainder is $r(x) = -17x + 13$.To make the polynomial divisible,we must add $-r(x) = -(-17x + 13) = 17x - 13$.

What must be added to the polynomial $p(x) = 6x^5 + 5x^4 + 11x^3 - 3x^2 + x + 1$ so that the resulting polynomial is exactly divisible by $3x^2 - 2x + 4$? (Hint: add $-r(x)$ to $p(x)$)

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