Answer:
[tex]\huge\boxed{C.\ f(x)=(x-6)(x+4)}[/tex]
Step-by-step explanation:
[tex]\text{If}\ x_1\ \text{and}\ x_2\ \text{are the zeros of quadratic function}\ f(x)=ax^2+bx+c,\\\\\text{then we can convert}\ f(x)\ \text{to the form}\ f(x)=a(x-x_1)(x-x_2).\\\\\text{We have the zeros}\ x_1=6\ \text{and}\ x_2=-4.\\\\\text{Substitute:}\\\\f(x)=a(x-6)(x-(-4))=a(x-6)(x+4)\\\\\text{For}\ a=1\ \text{we have answer}\ C.\ f(x)=(x-6)(x+4)[/tex]