Question

Is `121 + 11y + y^2` a perfect square trinomial and how do you factor it?

Answer 1

No, it's one of the factors of `y^3-11^3` a difference of cubes.

`y^3-11^3 = (y-11)(y^2+11y+11^2)`

If you were allowed complex coefficients then `y^3-11^3` factors as follows:

`y^3-11^3 = (y-11)(y-11omega)(y-11omega^2)`

where `omega` is called the primitive cube root of unity.

`omega = -1/2 + sqrt(3)/2i`