algebra problem please help?!?!?

Solve the following for all values of x: x^3 – x^2 – 18x + 18 = 0 ……..show work!

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  • x^3 - x^2 - 18x + 18 = 0

    can be factored by grouping

    x^2(x - 1) - 18(x - 1) = 0

    (x - 1)(x^2 - 18) = 0

    x = 1

    x^2 = 18

    x = ± 3√2

    Check

    QED

  • x^3 - x^2 - 18x + 18 = 0. Factor by grouping. Group the first two and last two terms separately.

    First group: x^3 - x^2. Factor out x^2 = x^2(x - 1). Come back to this later.

    Second group: 18x + 18. Factor out 18 = 18(x - 1). Put the two on one line.

    x^2(x - 1) - 18(x - 1) = 0. Note that you have (x - 1) that can be factored out. This will leave x^2 - 18 left.

    (x - 1)(x^2 - 18) = 0. Set each binomial = to 0.

    x - 1 = 0. x = 1. Answer.

    x^2 - 18 = 0. x^2 = 18. x = +/- sq rt 18. Sq rt 18 = (sq rt 9)(sq rt 2) = 3(sq rt 2). Thus,

    x = + 3(sq rt 2) and - 3(sq rt 2). Answer.

    Answers: x = 1, 3(sq rt 2), - 3(sq rt 2).

  • x³ – x² – 18x + 18 = 0

    (x³ – x²) – (18x – 18) = 0

    x²(x – 1) – 18(x – 1) = 0

    (x – 1)(x² – 18) = 0

    (x – 1) = 0

    x = 1

    (x² – 18) = 0

    x² = 18

    x = ± √18

    x = ± 3√2

  • x^3 - x^2 - 18x + 18 = 0

    (x^3 - x^2) - (18x - 18) = 0

    x^2(x - 1) - 18(x - 1) = 0

    (x - 1)(x^2 - 18) = 0

    x - 1 = 0

    x = 1

    x^2 - 18 = 0

    x^2 = 18

    x = ±√18

    x = ±√(2 * 3^2)

    x = ±3√2

    ∴ x = 1, ±3√2

  • solve by grouping

    x^3-x^2-18x+18

    x^2(x-1)-18(x-1)

    x^2-18 x-1

  • f ( 1 ) = 1 - 1 - 18 + 18 = 0

    Thus x - 1 is a factor

    Find other factors by synthetic division :-

    1 | 1______-1______-18______18

    _ |________1_______0______- 18

    _ | 1______ 0______- 18______ 0

    ( x - 1 ) ( x ² - 18 ) = 0

    x = 1 , x = ± √18

    x = 1 , x = ± 3 √ 2

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