The quotient when 3x^5-4x^4+5x^3-3x^2+6x-8 is | vedclass

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(B) To find the quotient,we perform polynomial long division of $p(x) = 3x^5-4x^4+5x^3-3x^2+6x-8$ by $t(x) = x^2+x-3$. Step $1$: Divide $3x^5$ by $x^2

Vedclass · Accepted Answer

$3x^3-7x^2+21x-45$

Vedclass · Answer

(B) To find the quotient,we perform polynomial long division of $p(x) = 3x^5-4x^4+5x^3-3x^2+6x-8$ by $t(x) = x^2+x-3$. Step $1$: Divide $3x^5$ by $x^2$ to get $3x^3$. Multiply $3x^3(x^2+x-3) = 3x^5+3x^4-9x^3$. Subtracting this from $p(x)$ gives $-7x^4+14x^3-3x^2+6x-8$. Step $2$: Divide $-7x^4$ by $x^2$ to get $-7x^2$. Multiply $-7x^2(x^2+x-3) = -7x^4-7x^3+21x^2$. Subtracting this gives $21x^3-24x^2+6x-8$. Step $3$: Divide $21x^3$ by $x^2$ to get $21x$. Multiply $21x(x^2+x-3) = 21x^3+21x^2-63x$. Subtracting this gives $-45x^2+69x-8$. Step $4$: Divide $-45x^2$ by $x^2$ to get $-45$. Multiply $-45(x^2+x-3) = -45x^2-45x+135$. Subtracting this gives $114x-143$. Thus,the quotient is $3x^3-7x^2+21x-45$.

The quotient when $3x^5-4x^4+5x^3-3x^2+6x-8$ is divided by $x^2+x-3$ is

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