Question

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 $ab+cd =$

• A. $p+2q$
• B. $p+2q-2$
• C. $2p+q$
• D. $2p+q-2$

Hint:

For polynomial long division, be systematic. At each step, focus only on the leading terms of the current dividend and the divisor to find the next term of the quotient. Then multiply and subtract carefully.

Answer 1

The Correct Option is B

We perform polynomial long division to find the quotient and remainder.
Divide $P(x) = 3x^5-6x^4+2x^3+4x^2-5x+8$ by $D(x) = x^2-2x+3$.
Step 1: Divide $3x^5$ by $x^2$ to get $3x^3$. $3x^3(x^2-2x+3) = 3x^5-6x^4+9x^3$. Subtract this from $P(x)$: $(3x^5-6x^4+2x^3) - (3x^5-6x^4+9x^3) = -7x^3$. Bring down the remaining terms: $-7x^3+4x^2-5x+8$.
Step 2: Divide $-7x^3$ by $x^2$ to get $-7x$. $-7x(x^2-2x+3) = -7x^3+14x^2-21x$. Subtract this: $(-7x^3+4x^2-5x) - (-7x^3+14x^2-21x) = -10x^2+16x$. Bring down the remaining term: $-10x^2+16x+8$.
Step 3: Divide $-10x^2$ by $x^2$ to get $-10$. $-10(x^2-2x+3) = -10x^2+20x-30$. Subtract this: $(-10x^2+16x+8) - (-10x^2+20x-30) = -4x+38$.
The division is complete.
The quotient is $Q(x) = 3x^3 - 7x - 10$. Comparing with $ax^3+bx^2+cx+d$, we get $a=3, b=0, c=-7, d=-10$.
The remainder is $R(x) = -4x+38$. Comparing with $px+q$, we get $p=-4, q=38$.
We need to calculate the value of $ab+cd$.
$ab+cd = (3)(0) + (-7)(-10) = 0 + 70 = 70$.
Now we check the options using $p=-4$ and $q=38$.
Option (A): $p+2q = -4 + 2(38) = -4 + 76 = 72$.
Option (B): $p+2q-2 = -4 + 2(38) - 2 = -4 + 76 - 2 = 70$.
Option (C): $2p+q = 2(-4)+38 = -8+38=30$.
Option (D): $2p+q-2 = 30-2=28$.
The value of $ab+cd$ is 70, which matches option (B).