Question

divide p(x)by q(x) and find the quotient and remainder
x^3-3x^2+4x-12=p(x);q(x)=(x+3)

Answer 1

Answer:

p(x)=x

3

−3x

2

+4x+2,g(x)=x−

Dividend =x

3

−3x

2

+4x+2

Divisor =x−1

Here, dividend and divisor both are in the standard form.

Now, on dividing p(x) by g(x) we get the following division process

Quotient =x

2

−2x+2

Remainder =4

Answer 2

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

Given that

$\rm :\longmapsto\:p(x) = {x}^{3} - {3x}^{2} + 4x - 12$

and

$\rm :\longmapsto\:q(x) = x + 3$

Using long division, we have

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{ \: \: \: \:\: {x}^{2} - 6x + 22\: \: \: \: \:}}}\\ {\underline{\sf{x + 3}}}& {\sf{\: {x}^{3}  -  { 3x}^{2} + 4x - 12 \:\:}} \\{\sf{}}& \underline{\sf{ -  {x}^{3} - 3 {x}^{2}   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:\:}} \\ {{\sf{}}}& {\sf{\: \:  \:  \:  \:  \:  \:  \:  \:  \: -6{x}^{2} + 4x - 12  \:  \:  \:  \:   \:  \:  \:  \:\:}} \\{\sf{}}& \underline{\sf{\:\: \:  \:  \:  \: \: 6 {x}^{2} + 18x  \:  \:  \:  \:  \:  \: \:\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \:  \:  \:  \:  \: \: \: \: \: \: \: 22x - 12  \:\:}} \\{\sf{}}& \underline{\sf{\: \:  \:  \:  \:  \:  - 22x  - 66\:\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \:  \: \:  \:  \:  \:  \:  \: \:  \:   - 78\:\:}}  \end{array}\end{gathered}\end{gathered}\end{gathered}$

Hence,

$\rm :\longmapsto\:Quotient = {x}^{2} - 6x + 22$

and

$\rm :\longmapsto\:Remainder = - \: 78$

Verification :-

We know,

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

Here,

$\rm :\longmapsto\:Quotient = {x}^{2} - 6x + 22$

$\rm :\longmapsto\:Remainder = - \: 78$

$\rm :\longmapsto\:Divisor = x + 3$

$\rm :\longmapsto\:Dividend= {x}^{3} - {3x}^{2} + 4x - 12$

Consider,

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

$\rm  \:  =  \: \:(x + 3)( {x}^{2} - 6x + 22) - 78$

$\rm  \:  =  \: \: {x}^{3} - {6x}^{2} + 22x + {3x}^{2} - 18x + 66 - 78$

$\rm  \:  =  \: \: {x}^{3 } - 3 {x}^{2} + 4x - 12$

$\rm  \:  =  \: \:Dividend$

Hence, Proved