## Introductory Algebra for College Students (7th Edition)

$\displaystyle \frac{1}{9}(3x+1)^{2}$
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: $... =\displaystyle \frac{1}{9}[(9x^{2}+3x)+(3x+1)]$ $... =\displaystyle \frac{1}{9}[3x(3x+1)+1(3x+1)]$ = $\displaystyle \frac{1}{9}(3x+1)(3x+1)$ = $\displaystyle \frac{1}{9}(3x+1)^{2}$