# Is 9z^2-3z+1 a perfect square trinomial, and how do you factor it?

Jun 12, 2015

No, it's not a perfect square trinomial. In fact it has no linear factors with real coefficients.

#### Explanation:

Perfect square trinomials are of the form:

${a}^{2} \pm 2 a b + {b}^{2} = {\left(a \pm b\right)}^{2}$

in this case we would be looking for $9 {z}^{2} - 6 z + 1 = {\left(3 z - 1\right)}^{2}$

Interestingly $9 {z}^{2} - 3 z + 1$ is one of the factors of ${\left(3 z\right)}^{3} + 1$

In general ${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

If you were allowed complex coefficients then you could write:

$27 {z}^{3} + 1 = {\left(3 z\right)}^{3} + {1}^{3}$

$= \left(3 z + 1\right) \left(9 {z}^{2} - 3 z + 1\right)$

$= \left(3 z + 1\right) \left(3 z + \omega\right) \left(3 z + {\omega}^{2}\right)$

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