Given z(x)=6x^3+bx^2-52x+15, z(2)=35, and z(-5)=0, algebracically determine all the zeros of z(x)

One way is to sub those values for x
when x=2, z(x)=35

35=6(2)³+b(2)²-52(2)+15
35=48+4b-104+15
35=4b-41
add 41 to both sides
76=4b
divide by 4
19=b

z(x)=6x³+19x²-52x+15

and z(-5)=0
that means (x+5) is a factor

divide using synthetic division or something
(6x³+19x²-52x+15)/(x+5)=6x²-11x+3
factor
6x²-11x+3=(2x-3)(3x-1)
set each to zero
2x-3=0
2x=3
x=3/2

3x-1=0
3x=1
x=1/3

th zeroes are at x=-5, 1/3, 2/3

127 millimeters is the answer