Given:
[tex]f(x)=4\mleft(x+7\mright)\mleft(x-5\mright)^2[/tex]
To find the real zero and its multiplicity:
Let the function equals to zero we get,
[tex]\begin{gathered} f(x)=0 \\ 4\mleft(x+7\mright)\mleft(x-5\mright)^2=0 \\ x+7=0 \\ \Rightarrow x=-7 \\ x-5=0 \\ \Rightarrow x=5 \end{gathered}[/tex]
Hence, the real zeros are -7 and 5.
The zero -7, multiplicity 1, crosses the x axis.
The zero 5, multiplicity 2, touches the x axis.
Hence, the correct option is the first one.
