Given the polynomial function \(f(x)=x^3-12x^2+41x-42.\)
The integer zeros should be divisors of free term -42.
The divisors of -42 are:
\(\pm 1, \pm 2, \pm 3, \pm 6, \pm 7, \pm 14, \pm 21, \pm 42.\)
If some of these number is zero of f(x), then f takes 0 value at this point.
Check them:
\(f(1)=1^3-12\cdot 1^2 + 41\cdot 1-42=1-12+41-42=-12
eq 0;\)
\(f(-1)=(-1)^3-12\cdot (-1)^2 + 41\cdot (-1)-42=-1-12-41-42=-96
eq 0;\)
\(f(2)=2^3-12\cdot 2^2 + 41\cdot 2-42=8-48+82-42=0;\)
\(f(-2)=(-2)^3-12\cdot (-2)^2 + 41\cdot (-2)-42=-8-48-82-42=-180
eq 0;\)
\(f(3)=3^3-12\cdot 3^2 + 41\cdot 3-42=27-108+123-42=0;\)
\(f(-3)=(-3)^3-12\cdot (-3)^2 + 41\cdot (-3)-42=-27-108-123-42=-300
eq 0;\)
\(f(6)=6^3-12\cdot 6^2 + 41\cdot 6-42=216-432+246-42=-12
eq 0;\)
\(f(-6)=(-6)^3-12\cdot (-6)^2 + 41\cdot (-6)-42=-216-432-246-42=-936
eq 0;\)
\(f(7)=7^3-12\cdot 7^2 + 41\cdot 7-42=343-588+287-42=0.\)
Since the third degree polynomial function may have only 3 zeros, then you can end this process and state that zeros are 2, 3 and 7, because f(2)=0, f(3)=0 and f(7)=0.
Answer: correct choice is C
