How do you know if #24a^2 + 26a + 9# is a perfect square trinomial and how do you factor it?

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Answer 1 · 2015-05-31T16:05:36

It isn't a perfect square trinomial. If it were then you would expect that the leading coefficient #24# would be square, which it isn't.

We can try matching the first and last coefficients with real numbers...

#(sqrt(24)a+3)^2 = 24a^2 + 6sqrt(24)a + 9#

Alternatively we can match the first two coefficients:

#(sqrt(24)a+(13sqrt(24))/24)^2#

#=24a^2+26a+169/24#

Worse, the discriminant is negative:

#Delta = b^2-4ac#

#= 26^2-(4xx24xx9) = 676 - 864 = -188#

This means that #24a^2+26a+9 = 0# has no real solutions. It has two distinct complex roots which are conjugates.

So #24a^2+26a+9# has no linear factors with real coefficients.