Question

If the remainders of the polynomial $f\left(x\right)$ when divided by $x-1$, $x-2$, $x-3$ are $1,3,7$ then the remainder of f(x) when divided by $(x-1)(x-2)(x-3)$ is

• A. $11$
• B. $x^2-x+1$
• C. $x^2+x+1$
• D. $12$

Answer 1

The correct option is B $x^2-x+1$

let us assume
$f\left(x\right)=\theta \left(x\right)(x-1)(x-2)(x-3)$
$+\gamma \left(x\right)$
where $\gamma \left(x\right)$ is the remainder
$since=orderofD\left(x\right)=3$
$\gamma \left(x\right)$ should be in the form of
$a{x}^2+bx+c$
$f\left(x\right)=\theta \left(x\right)(x-1)(x-2)(x-3)+a{x}^2+bx+c$
and $f\left(1\right)=1;f\left(2\right)=3;f\left(3\right)=7$
$1=a+b+C$
$3=4a+2b+c$
$7=ax+3b+c$
solving 3 equations we get
$a,b,c=1,-1,1$
So, the remainder is
$x^2-x+1$