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28 January, 01:32

Find all rational zeros of the function

h (x) = 6x^3-41x^2-8x+7

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Answers (2)
  1. 28 January, 01:39
    0
    -½, ⅓, 7

    Step-by-step explanation:

    The general formula for a third-degree polynomial is

    f (x) = ax³ + bx² + cx + d

    Your polynomial is

    h (x) = 6x³ - 41x² - 8x + 7

    a = 6; d = 7

    According to the rational root theorem, the possible rational roots are

    Factors of d/Factors of a

    Factors of d = ±1, ±7

    Factors of a = ±1, ±2, ±3, ±6

    Potential roots are x = ±1/1, ±1/2, ±1/3, ±1/6, ±7/1, ±7/2, ±7/3, ±7/6

    Putting them in order, we get the potential roots

    x = - 7, - ⁷/₂, - ⁷/₃, - ⁷/₆, - 1, - ½, - ⅓, - ⅙, ⅙, ⅓, ½, 1, ⁷/₆, ⁷/₃, ⁷/₂, 7

    Now, it's a matter of trial and error to find a zero.

    Let's try x = 7 by synthetic division.

    7|6 - 41 - 8 7

    | 42 7 - 7

    6 1 - 1 0

    So, x = 7 is a zero, and (6x³ - 41x² - 8x + 7) / (x - 7) = 6x² + x - 1

    Solve the quadratic.

    6x² + x - 1 = 0

    (3x - 1) (2x + 1) = 0

    3x - 1 = 0 2x + 1 = 0

    3x = 1 2x = - 1

    x = ⅓ x = - ½

    The zeroes of h (x) are - ½, ⅓, and 7.
  2. 28 January, 03:27
    0
    X = 7, 1/3,-1/2

    The zero are those above
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