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    Sketch the curve with equation:
    y = (x-1)^3 - 4(x-1)
    Showing the coordinates of the points where the curve crosses the x-axis

    _____

    Here's my working:
    (x-1)(x-1)(x-1) - 4(x-1)
    (x-1)(x^2-2x+1) - 4x+4
    x^3 - 2x^2 + x - x^2 + 2x - 1 - 4x + 4
    x^3 - 3x^2 +3x - 1 - 4x + 4
    x^3 - 3x^2 - x + 3

    I don't know what to do after i've worked out x^3 - 3x^2 - x + (unless i've done it wrong).
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    Factorise that into 3 brackets to get all the intercepts.
    As its +x^3 its drawn from the top right.
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    ah don't think i've been taught that, I know how to factorise it into a cubic if the +3 wasn't there.
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    (Original post by blankboi)
    ah don't think i've been taught that, I know how to factorise it into a cubic if the +3 wasn't there.
    can you get it into a quadratic and a single bracket.
    The number in the single bracket would have to multiply with C in the quadratic to give +3.
    If that makes any sense.
    Then factorise the quadratic.
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    (Original post by blankboi)
    Sketch the curve with equation:
    y = (x-1)^3 - 4(x-1)
    Showing the coordinates of the points where the curve crosses the x-axis

    _____

    Here's my working:
    (x-1)(x-1)(x-1) - 4(x-1)
    (x-1)(x^2-2x+1) - 4x+4
    x^3 - 2x^2 + x - x^2 + 2x - 1 - 4x + 4
    x^3 - 3x^2 +3x - 1 - 4x + 4
    x^3 - 3x^2 - x + 3

    I don't know what to do after i've worked out x^3 - 3x^2 - x + (unless i've done it wrong).
    There's a really easy way to factorise this.

    Observe that

    (x-1)^3 - 4(x-1) = (x-1)[(x-1)^2 - 4]

    If you now expand the contents of the square bracket, you'll get a quadratic that factorises

    That will leave you with all 3 points at which it crossss the x-axis


    Posted from TSR Mobile
 
 
 
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Updated: October 18, 2015
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