Answer
$p^2+9p+\frac{81}{4}=(p+\frac{9}{2})^2$
Work Step by Step
We add the square of half of the coefficent of $p$ so that the result is a perfect square trinomial.
Co-efficient of $p=9$
Half of 9 is $\frac{1}{2}×9=\frac{9}{2}$
Square of $\frac{9}{2}$ is $\frac{9}{2}×\frac{9}{2}=\frac{81}{4}$
We add $\frac{81}{4}$ to $p^2+9p$ to make it a perfect square trinomial.
Hence it becomes $p^2+9p+\frac{81}{4}$
Factored form-
$p^2+9p+\frac{81}{4}$
$= p^2+\frac{9}{2}p+\frac{9}{2}p +\frac{81}{4} $
$=p(p+\frac{9}{2})+\frac{9}{2} (p+\frac{9}{2})$ (Taking the common factors)
$=(p+\frac{9}{2}) (p+\frac{9}{2})$ ($(p+\frac{9}{2})$ is taken common from both the terms)
$=(p+\frac{9}{2})^2$
