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Question

(A) Perform polynomial long division of $3x^5-6x^4+2x^3+4x^2-5x+8$ by $x^2-2x+3$:
$1$. Divide $3x^5$ by $x^2$ to get $3x^3$. Multiply $3x^3(x^2-2x+3)

Vedclass · Accepted Answer

$70$

Vedclass · Answer

(A) Perform polynomial long division of $3x^5-6x^4+2x^3+4x^2-5x+8$ by $x^2-2x+3$:
$1$. Divide $3x^5$ by $x^2$ to get $3x^3$. Multiply $3x^3(x^2-2x+3) = 3x^5-6x^4+9x^3$. Subtracting this from the dividend gives $-7x^3+4x^2-5x+8$.
$2$. Divide $-7x^3$ by $x^2$ to get $-7x$. Multiply $-7x(x^2-2x+3) = -7x^3+14x^2-21x$. Subtracting this gives $-10x^2+16x+8$.
$3$. Divide $-10x^2$ by $x^2$ to get $-10$. Multiply $-10(x^2-2x+3) = -10x^2+20x-30$. Subtracting this gives $-4x+38$.
Thus,the quotient is $3x^3+0x^2-7x-10$ and the remainder is $-4x+38$.
Comparing with $ax^3+bx^2+cx+d$ and $px+q$,we have $a=3, b=0, c=-7, d=-10$ and $p=-4, q=38$.
Then $ab+cd = (3)(0) + (-7)(-10) = 0 + 70 = 70$.

If the quotient and remainder obtained when the expression $3x^5-6x^4+2x^3+4x^2-5x+8$ is divided by the expression $x^2-2x+3$ are $ax^3+bx^2+cx+d$ and $px+q$ respectively,then $ab+cd=$

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