Answer
$=x^2-3x+\frac{9}{4}$
$=(x-\frac{3}{2})^2$
Work Step by Step
The given expression is
$=x^2-3x$
Find the value of $(\frac{b}{2})^2$.
Substitute $-3$ for $b$.
$=(\frac{-3}{2})^2$
Simplify.
$=\frac{9}{4}$
Add $\frac{25}{4}$ to the given expression.
$=x^2-3x+\frac{9}{4}$
Write the the polynomial as $a^2-2ab+b^2$.
$=x^2-2(x)(\frac{3}{2})+(\frac{3}{2})^2$
Use perfect square trinomial pattern
$a^2-2ab+b^2=(a-b)^2$
We have $a=x$ and $b=\frac{3}{2}$.
$=(x-\frac{3}{2})^2$
