which set of values for x should be tested to determine the possible zeros of y = x^3 + 6x^2 - 10x + 35? (multiple choice)?

It's a cubic equation...

so possible zeros are when

0 = (x-a)(x-b)(x-c) 

==> 0 =x^3 -(a+b+c)x^2 + (ab + ac+bc)x -abc

so

-a*b*c = 35

factoring 35... one of the factors must be negative

So, looking at absolute values

|1| * |1| * |35|

|1| * |5| * |7|

so c) is not right -- 12 is not among the factors above

d) is not right -- one of the factors must be possibly negative

a) is out, since one of the factors must be 1 or -1

which leaves b) as the answer

x³ + 6x² - 10x + 35 = 0 → it's necessary to eliminate the term at the power 2

x³ + 6x² - 10x + 35 = 0 → let: x = z - 2

(z - 2)³ + 6.(z - 2)² - 10.(z - 2) + 35 = 0

[(z - 2)².(z - 2)] + 6.(z² - 4z + 4) - 10z + 20 + 35 = 0

[(z² - 4z + 4).(z - 2)] + 6z² - 24z + 24 - 10z + 55 = 0

[z³ - 2z² - 4z² + 8z + 4z - 8] + 6z² - 34z + 79 = 0

[z³ - 6z² + 12z - 8] + 6z² - 34z + 79 = 0

z³ - 6z² + 12z - 8 + 6z² - 34z + 79 = 0

z³ - 22z + 71 = 0 ← no term with power 2

z³ - 22z + 71 = 0 → let: z = u + v

(u + v)³ - 22.(u + v) + 71 = 0

[(u + v)².(u + v)] - 22.(u + v) + 71 = 0

[(u² + 2uv + v²).(u + v)] - 22.(u + v) + 71 = 0

[u³ + u²v + 2u²v + 2uv² + uv² + v³] - 22.(u + v) + 71 = 0

[u³ + v³ + 3u²v + 3uv²] - 22.(u + v) + 71 = 0

(u³ + v³) + (3u²v + 3uv²) - 22.(u + v) + 71 = 0

(u³ + v³) + 3uv.(u + v) - 22.(u + v) + 71 = 0 → you can factorize: (u + v)

(u³ + v³) + (u + v).(3uv - 22) + 71 = 0 → suppose that: (3uv - 22) = 0 ← equation (1)

(u³ + v³) + (u + v).(0) + 71 = 0

(u³ + v³) + 71= 0 ← equation (2)

You can get a system of 2 equations:

(1) : (3uv - 22) = 0

(1) : uv = 22/3

(1) : u³v³ = (22/3)³

(2) : (u³ + v³) + 71 = 0

(2) : u³ + v³ = - 71

Let: U = u³

Let: V = v³

You can get a new system of 2 equations:

(1) : UV = (22/3)³ ← this is the product P

(2) : U + V = - 71 ← this is the sum S

You know that the values U & V are the solutions of the following equation:

x² - Sx + P = 0 ← don’t confuse with the item x (initial equation)

x² + 71x + (22/3)³ = 0

Δ = 71² - [4 * (22/3)³]

Δ = 5041 - (42592/27)

Δ = 93515/27

x₁ = [- 71 + √(93515/27)]/2 ← this is U → recall: U = u³ → u = U^(1/3)

x₂ = [- 71 - √(93515/27)]/2 ← this is V → recall: V = v³ → v = V^(1/3)

Recall:

z = u + v

x = z - 2

x = u + v - 2

x = { [- 71 + √(93515/27)]/2 }^(1/3) + { [- 71 - √(93515/27)]/2 }^(1/3) - 2

x ≈ { - 1.8245772269254 } + { - 4.01919591295718 } - 2

x ≈ - 1.8245772269254 - 4.01919591295718 - 2

x ≈ - 7.84377313988259

I think they mean possible rational zeros, and  want you to use the rule that if b/a is a zero, then b is a factor of the leading coefficient while a is a factor of the constant term.  The factors of 35 are (plus or minus) 35, 7, 5 and 1 while the factors of 1 is just 1 so the possible rational zeros are ±35/1, ±7/1, V5/1 and ±1/1 so just try those.  IMO anyway.

So you must check both positive and negative factors 

and +/- 1 is always a factor.  

so for rational roots, you should check  (b) 

b) +- 1, +- 5, +- 7, +-35

You can try all positive and negative factors of 35 to find possible INTEGER zeroes. I'd start with the smallest ones (±1, then ±5 ...) because they're easier.

But until you find one there's no guarantee that there are any integer solutions. You should just type the equation into a site like https://www.wolframalpha.com/ or http://www.1728.org/cubic.htm to apply the cubic formula.