Answer
$\displaystyle \frac{1}{9}(3x+1)^{2}$
Work Step by Step
Write all coefficients with a common denominator:
$\displaystyle \frac{9}{9}x^{2}+\frac{6}{9}x+\frac{1}{9}$
...and factor out the gcf, $\displaystyle \frac{1}{9}...$
$...=\displaystyle \frac{1}{9}(9x^{2}+6x+1)$
... breaking $6x$ into $3x+3x$, we can factor in groups:
$... =\displaystyle \frac{1}{9}[(9x^{2}+3x)+(3x+1)]$
$... =\displaystyle \frac{1}{9}[3x(3x+1)+1(3x+1)]$
= $\displaystyle \frac{1}{9}(3x+1)(3x+1)$
= $\displaystyle \frac{1}{9}(3x+1)^{2}$
