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

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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+-2ab+b^2 = (a+-b)^2#

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

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

In general #a^3+b^3 = (a+b)(a^2-ab+b^2)#

If you were allowed complex coefficients then you could write:

#27z^3+1 = (3z)^3+1^3#

#= (3z+1)(9z^2-3z+1)#

#= (3z+1)(3z+omega)(3z+omega^2)#

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