Answer:
[tex]p=-7[/tex]
Step-by-step explanation:
Given [tex]f(x)=x^2+px+12[/tex], when [tex]f(x)=0[/tex], the right-hand side will break down to [tex](x+a)(x+b)[/tex] where [tex]a+b=p[/tex] and [tex]ab=12[/tex].
Because we know one of the zeroes is [tex]x=4[/tex], then either [tex]a=-4[/tex] or [tex]b=-4[/tex]. If we decide that [tex]a=-4[/tex], for example, and we expand [tex](x-4)(x+b)[/tex], we get [tex]x^2+(b-4)x-4b[/tex], which means [tex]b=-3[/tex] and [tex]p=-7[/tex].
We know this because [tex]-4(-3)=12[/tex] and [tex]-4+(-3)=-7[/tex].
