Question

Divided: x³-2x²-x+2 by x+1. Using long division method.​

Answer 1

ANSWER:

2

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Answer 2

$\large\underline{\sf{Solution-}}$

Given that

$\purple{\rm :\longmapsto\:Dividend = {x}^{3} - {2x}^{2} - x + 2}$

and

$\purple{\rm :\longmapsto\:Divisor = x + 1}$

So, By using Long Division Method, we have

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\:\: {x}^{2} - 3x + 2\:\:}}}\\ {\underline{\sf{x + 1}}}& {\sf{\: {x}^{3} - {2x}^{2} - x + 2 \:\:}} \\{\sf{}}& \underline{\sf{- {x}^{3} - {x}^{2} \: \: \: \: \: \: \: \: \: \: \:\:}} \\ {{\sf{}}}& {\sf{\: \: \: \: \: \: \: \: \: \: \: \: \: \: - 3{x}^{2} - x +2 \: \: \: \: \: \: \: \:\:}} \\{\sf{}}& \underline{\sf{\:\: \: \: \: \: \: \: \: \: \: \: 3{x}^{2} + 3x \: \: \: \: \: \: \:\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \: \: \: \: \: \: \: \: \: \: \: \: 2x + 2 \:\:}} \\{\sf{}}& \underline{\sf{\: \: \: \: \: \: \: \: \: \: \: \: - 2x - 2\:\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \: \: \: \: \: \: \: \: \: \: \: \: \: 0\:\:}} \end{array}\end{gathered}\end{gathered}\end{gathered}$

So,

$\bf\implies \:Remainder = 0$

Verification

$\purple{\rm :\longmapsto\:Dividend = {x}^{3} - {2x}^{2} - x + 2}$

$\purple{\rm :\longmapsto\:Divisor = x + 1}$

$\purple{\rm :\longmapsto\:Remainder = 0}$

$\purple{\rm :\longmapsto\:Quotient = {x}^{2} - 3x + 2}$

Now, Consider

$\rm :\longmapsto\:Divisor \times Quotient + Remainder$

$\rm \: = \: (x + 1)( {x}^{2} - 3x + 2) + 0$

$\rm \: = \: {x}^{3} - {3x}^{2} + 2x + {x}^{2} - 3x + 2$

$\rm \: = \: {x}^{3} - {2x}^{2} - x + 2$

$\rm \: = \: Dividend$

Hence, Verified