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