[tex]\large\underline{\sf{Solution-}}[/tex]
Given that
[tex] \purple{\rm :\longmapsto\:Dividend = {x}^{3} - {2x}^{2} - x + 2}[/tex]
and
[tex] \purple{\rm :\longmapsto\:Divisor = x + 1}[/tex]
So, By using Long Division Method, we have
[tex]\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\:\: {x}^{2} - 3x + 2\:\:}}}\\ {\underline{\sf{x + 1}}}& {\sf{\: {x}^{3} - {2x}^{2} - x + 2 \:\:}} \\{\sf{}}& \underline{\sf{- {x}^{3} - {x}^{2} \: \: \: \: \: \: \: \: \: \: \:\:}} \\ {{\sf{}}}& {\sf{\: \: \: \: \: \: \: \: \: \: \: \: \: \: - 3{x}^{2} - x +2 \: \: \: \: \: \: \: \:\:}} \\{\sf{}}& \underline{\sf{\:\: \: \: \: \: \: \: \: \: \: \: 3{x}^{2} + 3x \: \: \: \: \: \: \:\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \: \: \: \: \: \: \: \: \: \: \: \: 2x + 2 \:\:}} \\{\sf{}}& \underline{\sf{\: \: \: \: \: \: \: \: \: \: \: \: - 2x - 2\:\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \: \: \: \: \: \: \: \: \: \: \: \: \: 0\:\:}} \end{array}\end{gathered}\end{gathered}\end{gathered}[/tex]
So,
[tex]\bf\implies \:Remainder = 0[/tex]
Verification
[tex] \purple{\rm :\longmapsto\:Dividend = {x}^{3} - {2x}^{2} - x + 2}[/tex]
[tex] \purple{\rm :\longmapsto\:Divisor = x + 1}[/tex]
[tex] \purple{\rm :\longmapsto\:Remainder = 0}[/tex]
[tex] \purple{\rm :\longmapsto\:Quotient = {x}^{2} - 3x + 2}[/tex]
Now, Consider
[tex]\rm :\longmapsto\:Divisor \times Quotient + Remainder[/tex]
[tex]\rm \: = \: (x + 1)( {x}^{2} - 3x + 2) + 0[/tex]
[tex]\rm \: = \: {x}^{3} - {3x}^{2} + 2x + {x}^{2} - 3x + 2[/tex]
[tex]\rm \: = \: {x}^{3} - {2x}^{2} - x + 2[/tex]
[tex]\rm \: = \: Dividend[/tex]
Hence, Verified
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