Zeros are the x values which make the function equal to zero. Set it up as you would for a binomial with a constant multiplier "k" to account for the y-intercept (0, -5) given.
f(x) = k(x-2)(x-3)(x-5)
Use the y-intercept (0,-5) to solve for k.
-5 = k(0-2)(0-3)(0-5) -5 = -30k -5/-30 = k 1/6 = k
The cubic polynomial function is then ..
f(x) = (1/6)(x-2)(x-3)(x-5)
Linear factors are the linear (line) expressions you can factor out of the polynomial. They are (x-2), (x-3) and (x-5).