Question

The remainder when $x^4-y^4$ is divided by $x-y$ is

• A. $0$
• B. $x+y$
• C. $x^2-y^2$
• D. $2y^4$

Answer 1

Answer: (A)

$0$

Consider $x^4-y^4$

It can be expanded as $(x^2-y^2)(x^2+y^2)$

($\because$ $a^2-b^2=(a-b)(a+b)$)

It can be further expanded as $(x-y)(x+y)(x^2+y^2)$

Now,we can observe that $x-y$ is a factor of $x^4-y^4$

Hence, on dividing $x^4-y^4$ with $x-y$ we get a remainder of Zero.