Answer:
A) Notice that:
[tex]\begin{gathered} x^2-12x+35=x^2+(-5-7)x+(-5)(-7) \\ =x^2-5x-7x+(-5)(-7)=x(x-5)-7(x-5) \\ =(x-7)(x-5)\text{.} \end{gathered}[/tex]
Therefore:
[tex]x^2-12x+35=0\text{ if and only if x=7 or x=5.}[/tex]
B) The solutions from part A represent the x-coordinates of the x-intercepts of the graph of the function
[tex]f(x)=x^2-12x+35.[/tex]
