Question:

On dividing $x^{3} - 3x^{2} + x + 2$ by a polynomial $g(x)$, the quotient and remainder were $x - 2$ and $-2x + 4$ respectively. Find $g(x)$.

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With known quotient and remainder, recover the divisor via $g(x)=\dfrac{f(x)-r(x)}{q(x)}$.

Updated On: Jul 16, 2026

$x^{2}+2x+1$
$x^{2}-2x+1$
$x+x+1$
$x^{2}-x+1$

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The Correct Option is D

Approach Solution - 1

Dividend $f(x)=x^{3}-3x^{2}+x+2$, quotient $q(x)=x-2$, remainder $r(x)=-2x+4$. Use $f(x)=g(x)q(x)+r(x)$: \[ g(x)=\frac{f(x)-r(x)}{q(x)} =\frac{x^{3}-3x^{2}+x+2-(-2x+4)}{x-2} =\frac{x^{3}-3x^{2}+3x-2}{x-2}. \] Divide: $(x-2)(x^{2}-x+1)=x^{3}-3x^{2}+3x-2$. Thus $g(x)=\boxed{x^{2}-x+1}$.

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Approach Solution -2

Instead of directly dividing $f(x)-r(x)$ by $q(x)$, we can assume the form of $g(x)$ and match coefficients, then check each option.

Option A ($x^2+2x+1$): Since $f(x)=g(x)q(x)+r(x)$ with $q(x)=x-2$ of degree $1$ and $f(x)$ of degree $3$, $g(x)$ must be a monic quadratic; let $g(x)=x^2+bx+c$. Then \[ (x^2+bx+c)(x-2)+(-2x+4)=x^3+(b-2)x^2+(c-2b-2)x+(4-2c). \] Matching with $f(x)=x^3-3x^2+x+2$: from the $x^2$ term, $b-2=-3\Rightarrow b=-1$; from the constant term, $4-2c=2\Rightarrow c=1$. This gives $g(x)=x^2-x+1$, not $x^2+2x+1$, so this option is incorrect.
Option B ($x^2-2x+1$): This does not match $b=-1,\,c=1$ found above, so it is incorrect.
Option C ($x+x+1$): This is not even a valid quadratic expression (it simplifies to $2x+1$, degree $1$), so it cannot be $g(x)$, which must be degree $2$.
Option D ($x^2-x+1$): This matches $b=-1$, $c=1$ exactly, as found above. Checking the $x$-coefficient: $c-2b-2=1-(-2)-2=1$, which matches the coefficient of $x$ in $f(x)$, confirming consistency.

Matching coefficients after assuming a monic quadratic form for $g(x)$ confirms $g(x)=x^2-x+1$.

Hence, the correct answer is option D: $x^2-x+1$.