\(\displaystyle{y}={3}{x}^{{{2}}}\)

Pick some factor to multiply top and bottom

\(\displaystyle{y}={3}{x}^{{{2}}}\cdot{{\lbrace}}{\frac{{{x}+{2}}}{{{x}+{2}}}}={\frac{{{3}{x}^{{{3}}}+{6}{x}^{{{2}}}}}{{{x}+{2}}}}\}\)

This adds a "hole" at \(x=-2\) simce that would be undefined, but otherwise behaves exactly like \(\displaystyle{y}={3}{x}^{{{2}}}\) because of course the factors cancel out.

2. Pick another factor to multiply top and bottom

\(\displaystyle{y}={3}{x}^{{{2}}} \cdot {{\lbrace}}{\frac{{{x}-{5}}}{{{x}-{5}}}}={\frac{{{3}{x}^{{{3}}}-{15}{x}^{{{2}}}}}{{{x}-{5}}}}\}\)

"Hole" at x=5, but otherwise the same as \(\displaystyle{y}={3}{x}^{{{2}}}\).

A simpler option could be to just add some constant to \(\displaystyle{3}{x}^{{{2}}}\)