Question:
If \(5\) is the remainder when
\[
2x^5+kx^4+5x^3-3x^2+2x-1
\]
is divided by
\[
x^2+x+1,
\]
then the quotient is:
If \(5\) is the remainder when
\[
2x^5+kx^4+5x^3-3x^2+2x-1
\]
is divided by
\[
x^2+x+1,
\]
then the quotient is:
Show Hint
Whenever the divisor is \(x^2+x+1\), remember the identities
\[
\omega^3=1
\quad\text{and}\quad
1+\omega+\omega^2=0.
\]
Using roots of unity often avoids lengthy polynomial division and helps determine unknown coefficients quickly.
Updated On: Jun 10, 2026
- \[ 2x^3-x^2+10x+4 \]
- \[ 2x^3-5x^2+8x-6 \]
- \[ 2x^3-5x^2+10x+4 \]
- \[ 2x^3-x^2+8x-6 \]
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Concept:
When a polynomial \(P(x)\) is divided by another polynomial \(D(x)\), we use
\[
P(x)=D(x)\cdot Q(x)+R(x),
\]
where \(Q(x)\) is the quotient and \(R(x)\) is the remainder.
Since the divisor
\[
x^2+x+1
\]
is of degree \(2\), the remainder must be of degree less than \(2\).
The question states that the remainder is \(5\), therefore
\[
R(x)=5.
\]
Hence
\[
2x^5+kx^4+5x^3-3x^2+2x-1
=
(x^2+x+1)Q(x)+5.
\]
We shall determine \(k\) and then identify the quotient.
Step 1: Use the remainder condition Let \[ P(x)=2x^5+kx^4+5x^3-3x^2+2x-1. \] Since the remainder is \(5\), \[ P(x)-5 \] must be exactly divisible by \[ x^2+x+1. \] Thus \[ P(x)-5 = 2x^5+kx^4+5x^3-3x^2+2x-6. \]
Step 2: Use the factor relation Since \[ x^2+x+1=0 \] has roots \[ \omega,\omega^2, \] where \[ \omega^3=1, \qquad 1+\omega+\omega^2=0, \] we must have \[ P(\omega)-5=0. \] Using \[ \omega^3=1, \qquad \omega^4=\omega, \qquad \omega^5=\omega^2, \] we get \[ 2\omega^2+k\omega+5-3\omega^2+2\omega-6=0. \] Simplifying, \[ (k+2)\omega-\omega^2-1=0. \] Since \[ 1+\omega+\omega^2=0, \] we have \[ -\omega^2-1=\omega. \] Hence \[ (k+2)\omega+\omega=0. \] \[ (k+3)\omega=0. \] Therefore, \[ k=-3. \] \[ \boxed{k=-3} \]
Step 3: Substitute \(k=-3\) The polynomial becomes \[ 2x^5-3x^4+5x^3-3x^2+2x-1. \] Since remainder is \(5\), \[ P(x)-5 = 2x^5-3x^4+5x^3-3x^2+2x-6. \] Now divide by \[ x^2+x+1. \]
Step 4: Polynomial division Dividing \[ 2x^5-3x^4+5x^3-3x^2+2x-6 \] by \[ x^2+x+1, \] the quotient obtained is \[ 2x^3-5x^2+8x-6. \] Verification: \[ (x^2+x+1)(2x^3-5x^2+8x-6) \] \[ = 2x^5-3x^4+5x^3-3x^2+2x-6. \] Adding the remainder \(5\), \[ 2x^5-3x^4+5x^3-3x^2+2x-1. \] which is exactly the original polynomial. Hence the quotient is \[ \boxed{2x^3-5x^2+8x-6}. \] Therefore the correct option is \[ \boxed{(B)}. \]
Step 1: Use the remainder condition Let \[ P(x)=2x^5+kx^4+5x^3-3x^2+2x-1. \] Since the remainder is \(5\), \[ P(x)-5 \] must be exactly divisible by \[ x^2+x+1. \] Thus \[ P(x)-5 = 2x^5+kx^4+5x^3-3x^2+2x-6. \]
Step 2: Use the factor relation Since \[ x^2+x+1=0 \] has roots \[ \omega,\omega^2, \] where \[ \omega^3=1, \qquad 1+\omega+\omega^2=0, \] we must have \[ P(\omega)-5=0. \] Using \[ \omega^3=1, \qquad \omega^4=\omega, \qquad \omega^5=\omega^2, \] we get \[ 2\omega^2+k\omega+5-3\omega^2+2\omega-6=0. \] Simplifying, \[ (k+2)\omega-\omega^2-1=0. \] Since \[ 1+\omega+\omega^2=0, \] we have \[ -\omega^2-1=\omega. \] Hence \[ (k+2)\omega+\omega=0. \] \[ (k+3)\omega=0. \] Therefore, \[ k=-3. \] \[ \boxed{k=-3} \]
Step 3: Substitute \(k=-3\) The polynomial becomes \[ 2x^5-3x^4+5x^3-3x^2+2x-1. \] Since remainder is \(5\), \[ P(x)-5 = 2x^5-3x^4+5x^3-3x^2+2x-6. \] Now divide by \[ x^2+x+1. \]
Step 4: Polynomial division Dividing \[ 2x^5-3x^4+5x^3-3x^2+2x-6 \] by \[ x^2+x+1, \] the quotient obtained is \[ 2x^3-5x^2+8x-6. \] Verification: \[ (x^2+x+1)(2x^3-5x^2+8x-6) \] \[ = 2x^5-3x^4+5x^3-3x^2+2x-6. \] Adding the remainder \(5\), \[ 2x^5-3x^4+5x^3-3x^2+2x-1. \] which is exactly the original polynomial. Hence the quotient is \[ \boxed{2x^3-5x^2+8x-6}. \] Therefore the correct option is \[ \boxed{(B)}. \]
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