Question

Factor the expression completely over the complex numbers.

y^4-16

Answer 1

We were given;


`y^4 -16` to factor over the complex numbers.

We can rewrite the given expression to look like difference of two squares,


`y^4 -16=(y^2)^(2) -4^2`


`y^4 -16=(y^2+4)(y^2-4)`


`y^4 -16=(y^2--4)(y^2-4)`

We can still rewrite as,


`y^4 -16=(y^2--2^2)(y^2-2^2)`

We can still rewrite as


`y^4 -16=(y^2-2^2*-1)(y^2-2^2)`


`y^4 -16=(y^2-2^2(-1))(y^2-2^2)`

Applying difference of two squares again, we have;


`y^4 -16=(y+2√(-1))(y-2√(-1))(y+2)(y-2)`

Note that, in complex numbers


`√(-1)=i`


`y^4 -16=(y+2i)(y-2i)(y+2)(y-2)`