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

1 Answer
May 30, 2015

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#