Yahoo Answers is shutting down on 4 May 2021 (Eastern Time) and the Yahoo Answers website is now in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

Anonymous
Anonymous asked in Science & MathematicsMathematics · 7 days ago

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

a) +- 5, +-7, +- 35

b) +- 1, +- 5, +-  7, +-35

c) 1, 5, 7, 12, and 35

d) 1, 5, 7, and 35

5 Answers

Relevance
  • 7 days ago
    Favourite answer

    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

     

  • 7 days ago

    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

  • 7 days ago

    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.

  • ?
    Lv 7
    7 days ago

    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

  • ?
    Lv 7
    7 days ago

    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.

Still have questions? Get answers by asking now.