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)

Answer 1

$\large\sf\underline{Given\::}$

• $\sf\:x^{3}-3x^{2}+x+2$ is divided by $\sf\:g(x)$

• Quotient = $\sf\:(x-2)$

• Remainder = $\sf\:(-2x+4)$

‎

$\large\sf\underline{To\:find\::}$

• Value of $\sf\:g(x)$.

‎

$\large\sf\underline{Concept~to~be~used\::}$

$\small{\underline{\boxed{\mathrm{Dividend~=~Divisor~\times~Quotient~+~Remainder}}}}$

‎

• Dividend :- The number to be divided.

‎

• Divisor :- The number which divides the other number.

‎

• Quotient :- Number that we get as a result after dividing dividend by a divisor.

‎

• Remainder :- After dividing a dividend by a divisor if some number is left which can't be divided further, that number is known as remainder.

‎

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

Using,

$\sf\:Dividend~=~Divisor~\times~Quotient~+~Remainder$

$\sf\dashrightarrow\:x^{3}-3x^{2}+x+2=g(x) \times (x-2) +(-2x+4)$

$\sf\dashrightarrow\:x^{3}-3x^{2}+x+2=g(x) \times (x-2) - 2x+4$

‎

• Transposing ( -2x+4 ) to LHS it becomes ( 2x-4 )

‎

$\sf\dashrightarrow\:x^{3}-3x^{2}+x+2+2x-4=g(x) \times (x-2)$

‎

• Arranging the terms in LHS for calculation

‎

$\sf\dashrightarrow\:x^{3}-3x^{2}+x+2x+2-4=g(x) \times (x-2)$

$\sf\dashrightarrow\:x^{3}-3x^{2}+3x-2=g(x) \times (x-2)$

‎

• Transposing ( x-2 ) from RHS to LHS it goes to the denominator

‎

$\sf\dashrightarrow\:\frac{x^{3}-3x^{2}+3x-2}{x-2}=g(x)$

$\sf\:oR\:g(x)~=~\frac{x^{3}-3x^{2}+3x-2}{x-2}$

‎

• Now let's proceed with division in order to get simplified value of g(x)

‎

$\sf\:x-2)\cancel{x^{3}}-3x^{2}+3x-2(x^{2}-x+1$

$\sf\:~~~~~~~~~~\cancel{x^{3}}-2x^{2}$

$\sf\:~~~~~~~~~~---------$

$\sf\:~~~~~~~~~~~~~~~~~~~~\cancel{x^{2}}+3x-2$

$\sf\:~~~~~~~~~~~~~~~~~~~~\cancel{x^{2}}+2x$

$\sf\:~~~~~~~~~~---------$

$\sf\:~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\cancel{x-2}$

$\sf\:~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\cancel{x-2}$

$\sf\:~~~~~~~~~~---------$

$\sf\:~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~0$

‎

Therefore, we got the quotient as $\tt\:x^{2}-x+1$

Henceforth the required value of :

$\dag\:\underline{\sf~g(x)~=~x^{2}-x+1}$

‎________________

Note :- Scroll left to right to view the answer properly !~

Answer 2

$[tex]\large\sf\underline{Given\::}$

$\sf\:x^{3}-3x^{2}+x+2$ is divided by $\sf\:g(x)$

Quotient = $\sf\:(x-2)$

Remainder = $\sf\:(-2x+4)$

‎

$\large\sf\underline{To\:find\::}$

Value of $\sf\:g(x)$.

‎

$\large\sf\underline{Concept~to~be~used\::}$

$\small{\underline{\boxed{\mathrm{Dividend~=~Divisor~\times~Quotient~+~Remainder}}}}$

‎

Dividend :- The number to be divided.

‎

Divisor :- The number which divides the other number.

‎

Quotient :- Number that we get as a result after dividing dividend by a divisor.

‎

Remainder :- After dividing a dividend by a divisor if some number is left which can't be divided further, that number is known as remainder.

‎

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

Using,

$\sf\:Dividend~=~Divisor~\times~Quotient~+~Remainder$

$\sf\dashrightarrow\:x^{3}-3x^{2}+x+2=g(x) \times (x-2) +(-2x+4)$

$\sf\dashrightarrow\:x^{3}-3x^{2}+x+2=g(x) \times (x-2) - 2x+4$

‎

Transposing ( -2x+4 ) to LHS it becomes ( 2x-4 )

‎

$\sf\dashrightarrow\:x^{3}-3x^{2}+x+2+2x-4=g(x) \times (x-2)$

‎

Arranging the terms in LHS for calculation

‎

$\sf\dashrightarrow\:x^{3}-3x^{2}+x+2x+2-4=g(x) \times (x-2)$

$\sf\dashrightarrow\:x^{3}-3x^{2}+3x-2=g(x) \times (x-2)$

‎

Transposing ( x-2 ) from RHS to LHS it goes to the denominator

‎

$\sf\dashrightarrow\:\frac{x^{3}-3x^{2}+3x-2}{x-2}=g(x)$

$\sf\:oR\:g(x)~=~\frac{x^{3}-3x^{2}+3x-2}{x-2}$

‎

Now let's proceed with division in order to get simplified value of g(x)

‎

$\sf\:x-2)\cancel{x^{3}}-3x^{2}+3x-2(x^{2}-x+1$

$\sf\:~~~~~~~~~~\cancel{x^{3}}-2x^{2}$

$\sf\:~~~~~~~~~~---------$

$\sf\:~~~~~~~~~~~~~~~~~~~~\cancel{x^{2}}+3x-2$

$\sf\:~~~~~~~~~~~~~~~~~~~~\cancel{x^{2}}+2x$

$\sf\:~~~~~~~~~~---------$

$\sf\:~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\cancel{x-2}$

$\sf\:~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\cancel{x-2}$

$\sf\:~~~~~~~~~~---------$

$\sf\:~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~0$

‎

Therefore, we got the quotient as $\tt\:x^{2}-x+1$

Henceforth the required value of :

$\dag\:\underline{\sf~g(x)~=~x^{2}-x+1}$

‎________________

Note :- Scroll left to right to view the answer properly !~[/tex]