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9 October, 06:21

Factorise:-

x^3 - 6x^2 + 11x - 6

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Answers (2)
  1. 9 October, 08:51
    0
    (x - 1) (x - 2) (x - 3).

    Step-by-step explanation:

    f (x) = x^3 - 6x^2 + 11x - 6

    f (1) = 1 - 6 + 11 - 6 = 12 - 12

    = 0 so by the Factor Theorem (x - 1) is a factor.

    Also, by the Rational Root theorem, as the last term is - 6 and the leading coefficient is 1 some of - 1, 2,-2, 3 - 3, 6,-6 might be zeroes of the function.

    f (-1) = - 1 + 6 - 11 - 6 = - 12 so - 1 is not a zero and (x + 1) is not a factor.

    f (2) = 8 - 24 + 22 - 6

    = 30 - 30 = 0 so (x - 2) is also a factor)

    Since the last term is - 6 the last factor must be (x - 3)

    =Checking: f (3) = 27 - 6 (9) + 33 - 6

    = 27 - 54 + 33 - 6

    = 60-60 = 0.

    So the factors are (x - 1) (x - 2) (x - 3).
  2. 9 October, 09:04
    0
    (x - 1) (x - 2) (x - 3)

    Step-by-step explanation:

    Note the sum of the coefficients

    1 - 6 + 11 - 6 = 0

    hence x = 1 is a root and (x - 1) is a factor

    dividing x³ - 6x² + 11x - 6 by (x - 1) gives

    (x - 1) (x² - 5x + 6)

    To factor the quadratic

    Consider the factors of + 6 which sum to give - 5

    The factors are - 2 and - 3, since

    - 2 * - 3 = 6 and - 2 - 3 = - 5, hence

    x² - 5x + 6 = (x - 2) (x - 3) and

    x³ - 6x² + 11x - 6 = (x - 1) (x - 2) (x - 3)
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