Question

the remainder when x^4-y^4 is divided by x-y is------------?

Answer 1

x⁴-y⁴=(x²+y²)(x²-y²)=(x²+y²)(x+y)(x-y). Hence when ,this term is divided by (x-y) it will be perfectly divisible as it contains the term (x-y). Therefore the remainder will be zero .

Answer 2

The remainder is $(x^2+y^2)(x+y)$.

Step-by-step explanation:

We have,

$\dfrac{x^4-y^4}{x-y}$

To find, the remainder of $\dfrac{x^4-y^4}{x-y}=?$

∴$x^4-y^4$

$=(x^2)^2-(y^2)^2$

$=(x^2+y^2)(x^2-y^2)$

Using algebraic identity,

$a^{2}-b^{2}=(a+b)(a-b)$

$=(x^2+y^2)(x+y)(x-y)$

∴$\dfrac{x^4-y^4}{x-y}$

$=\dfrac{(x^2+y^2)(x+y)(x-y)}{x-y}$

$=(x^2+y^2)(x+y)$

Hence, the remainder is $(x^2+y^2)(x+y)$.