well, we know is a quadratic, namely of degree 2, so it has two solutions and namely two factors, and whose only zero is 7, well, we can do that by giving that only zero a multiplicity of 2, so we get two of the same zero.
[tex]\begin{cases} x=7\implies &x-7=0\\ x=7\implies &x-7=0 \end{cases}\qquad \implies \qquad \begin{array}{llll} \stackrel{\textit{multiplicity of 2}}{(x-7)(x-7)}=\stackrel{0}{y} \\\\\\ x^2-14x+49=y \end{array}[/tex]
