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

Question

1867 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-12T18:06:46

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

Perfect square trinomials are of the form:

$a^{2} \pm 2 a b + b^{2} = {(a \pm b)}^{2}$ $a^2+-2ab+b^2 = (a+-b)^2$

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

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

In general $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ $a^3+b^3 = (a+b)(a^2-ab+b^2)$

If you were allowed complex coefficients then you could write:

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

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

$= (3 z + 1) (3 z + ω) (3 z + ω^{2})$ $= (3z+1)(3z+omega)(3z+omega^2)$

where $ω = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ $omega = -1/2 + sqrt(3)/2i$

Is 9z^2-3z+19z2−3z+19z^2-3z+1 a perfect square trinomial, and how do you factor it?