# What should be added to 6x^{4}-x^{3}+3x^{2}+4x-2 so that the result is exactly divisible by 3x^{2}-2x+4 ?

Since ,

Dividend = Divisor * Quotient +Remainder

6x^{4}-x^{3}+3x^{2}+4x-2=(3x^{2}-2x+4)*(2x^{2}+x-1)+(-2x+2)

So to get remainder ‘0’, we should append “ –(-2x+2)”; that is, we should add ‘2x-2’ to the dividend to get the remainder ‘0’

Actually in such type of sums , first we do regular division , then we see the remainder , opposite of the remainder should be always added to make the division with remainder zero .