Answer
${{\left( 3xy-5 \right)}^{2}}$
Work Step by Step
$9{{x}^{2}}{{y}^{2}}-30xy+25$
Rewrite the equation as shown below,
$9{{x}^{2}}{{y}^{2}}-30xy+25={{\left( 3xy \right)}^{2}}-2\cdot \left( 3xy \right)\cdot \left( 5 \right)+{{\left( 5 \right)}^{2}}$
The sign of the middle term is negative, so:
${{A}^{2}}-2AB+{{B}^{2}}={{\left( A-B \right)}^{2}}$
$\begin{align}
& 9{{x}^{2}}{{y}^{2}}-30xy+25={{\left( 3xy \right)}^{2}}-2\cdot \left( 3xy \right)\cdot \left( 5 \right)+{{\left( 5 \right)}^{2}} \\
& ={{\left( 3xy-5 \right)}^{2}}
\end{align}$
Check:
Multiply the factors and simplify using the distributive law as shown below:
$\begin{align}
& {{\left( 3xy-5 \right)}^{2}}=\left( 3xy-5 \right)\left( 3xy-5 \right) \\
& =3xy\left( 3xy-5 \right)-5\left( 3xy-5 \right) \\
& ={{\left( 3xy \right)}^{2}}-15xy-15xy+{{\left( 5 \right)}^{2}} \\
& =9{{x}^{2}}{{y}^{2}}-30xy+25
\end{align}$
Therefore, the factors of the polynomial $9{{x}^{2}}{{y}^{2}}-30xy+25$ are ${{\left( 3xy-5 \right)}^{2}}$ .
