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

May 31, 2015

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...

${\left(\sqrt{24} a + 3\right)}^{2} = 24 {a}^{2} + 6 \sqrt{24} a + 9$

Alternatively we can match the first two coefficients:

${\left(\sqrt{24} a + \frac{13 \sqrt{24}}{24}\right)}^{2}$

$= 24 {a}^{2} + 26 a + \frac{169}{24}$

Worse, the discriminant is negative:

$\Delta = {b}^{2} - 4 a c$

$= {26}^{2} - \left(4 \times 24 \times 9\right) = 676 - 864 = - 188$

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

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