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

May 30, 2015

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

${y}^{3} - {11}^{3} = \left(y - 11\right) \left({y}^{2} + 11 y + {11}^{2}\right)$

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

${y}^{3} - {11}^{3} = \left(y - 11\right) \left(y - 11 \omega\right) \left(y - 11 {\omega}^{2}\right)$

where $\omega$ is called the primitive cube root of unity.

$\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$