Question

If $x^5$ is divided by $x^2-4x+3$, then find its remainder.

• A. $121x+120$
• B. $121x-120$
• C. $120x+121$
• D. $120x-121$

Answer 1

The correct option is B $121x-120$

Given that:

$p\left(x\right)=x^5,g\left(x\right)=x^2-4x+3$

Solution:

We know that,

$p\left(x\right)=g\left(x\right).q\left(x\right)+r\left(x\right)$

Let us suppose $r\left(x\right)=Ax+B$

or. $x^5=(x^2-4x+3)\times q\left(x\right)+Ax+B$

or, $x^5=(x-1)(x-3)\times q\left(x\right)+Ax+B$

Put $x=1$ we get,

or, $1^5=0\times q\left(x\right)+A+B$

or, $A+B=1$ $.................\left(i\right)$

Put $x=3$ we get,

or, $3^5=0\times q\left(x\right)+3A+B$

or, $3A+B=243$ $...............\left(ii\right)$

Subtracting eqn.(i) from (ii) we get,

$2A=242$

or, $A=121$

And $B=1-121=-120$

Hence, Remainder $r\left(x\right)=121x-120$

Therefore, B is the correct option.