Question

(x³+6x²-x-18)÷(x+2)
please give the proper solution.​

Answer 1

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

Given expression is

$\rm :\longmapsto\:\dfrac{ {x}^{3} + {6x}^{2} - x - 18}{x + 2}$

So, here

$\rm :\longmapsto\:Dividend : {x}^{3} + 6 {x}^{2} - x - 18$

$\rm :\longmapsto\:Divisor : x + 2$

So, using Long Division Method, we have

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\:\: {x}^{2} + 4x - 9\:\:}}}\\ {\underline{\sf{x + 2}}}& {\sf{\: {x}^{3} + {6x}^{2} - x - 18\:\:}} \\{\sf{}}& \underline{\sf{\:\:   -  {x}^{3} - 2{x}^{2} \:   \: \: \: \: \:  \:  \:  \:  \:  \:  \:  \:  \: \:\:}} \\ {{\sf{}}}& {\sf{\: \:  \:  \:  \: 4{x}^{2} - x - 18}} \\{\sf{}}& \underline{\sf{\:\: \:  \:   - 4 {x}^{2}  - 8x  \:  \:  \:  \: \:\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \:  \:  \:  \:  \:    \:  \:  -9x - 18  \:\:}} \\{\sf{}}& \underline{\sf{\: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  9x + 18\:\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \:  \:  \: \: \: \: \: \: \:  \:  0\:\:}}  \end{array}\end{gathered}\end{gathered}\end{gathered}$

So, it means

$\rm \implies\:Quotient : {x}^{2} + 4x - 9$

Hence,

$\rm :\longmapsto\:\boxed{\tt{ \dfrac{ {x}^{3} + {6x}^{2} - x - 18}{x + 2} = {x}^{2} + 4x - 9}}$

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Verification :-

We have,

$\rm :\longmapsto\:Divisor : x + 2$

and

$\rm \implies\:Quotient : {x}^{2} + 4x - 9$

Now, Consider

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

$\rm \:  =  \: (x + 2)( {x}^{2} + 4x - 9) + 0$

$\rm \:  =  \: {x}^{3} + {4x}^{2} - 9x + {2x}^{2} + 8x - 18$

$\rm \:  =  \: {x}^{3} + {6x}^{2} - x - 18$

$\rm \:  =  \: Dividend$

Hence, Verified