Question

The remainder when x^5 +kx² is divided by (x-1)(x-2)(x-3) contains no term in x² .then find the value of k.

Answer 1

we see (x-1)(x-2)(x-3) gain x^3
it means when we divided nomenator to denominator x^2 , x and constant outcomes .
but question ask
x^2 doesn't contains after that
hence ,
x^5+kx^2/(x-1)(x-2)(x-3)=kx+c(let )
where k and c are constant
now,
x^5+kx^2=(x-1)(x-2)(x-3)(kx+c)
={x^2-3x+2}{kx^2+cx-3kx-3c}
={x^2-3x+2}{kx^2 +(c-k) x-3c}
=kx^4+(c-k) x^3-3cx^2-3kx^3-3(c-k) x^2+9cx+2kx^2+2 (c-k)x-6c

now compare LHS and RHS
left side x^4 doesn't exist so,
right side coefficient of x^4=0
hence k=0