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Factorising to form linear factor and quadratic

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    How do i factorise a cubic? the equation is:

    2x^3+9x^2+11x-8

    i am giving (2x-1) is a factor.

    so far I have (2x-1)(x^2+5x.....)
    any tips on a how to in an exam would be helpful too

    Thanks in advance
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    Yo, by far my favourite method for polynomials of small degree is something which I've forgotten the name of! Comparison of factors or something?

    Anyway, we write a general quadratic and note (2x-1)(ax^2+bx+c), then we note that the only way to get an x^3 is by the 2x * ax^2, so 2ax^3 = 2x^3 implies a = 1, then we note that the only way to get a constant is -1*c, so -c = -8, c= 8, this leaves the trickier b, which is like, 2x*c - bx right? Soooo
    x(16-b) = 11x, aha! b is 5, just as you suspected.

    This idea can be extended to polynomials of n degree, make a general polynomial of n-1 degree, but with big n this can get messy.

    Edit: Seriously, what is that called, it's killing me.
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    I'm assuming you know how to do polynomial division? If you do, divide your cubic by 2x-1
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    (Original post by Stickyelmo)
    How do i factorise a cubic? the equation is:

    2x^3+9x^2+11x-8

    i am giving (2x-1) is a factor.

    so far I have (2x-1)(x^2+5x.....)
    any tips on a how to in an exam would be helpful too

    Thanks in advance
     f(x) = 2x^3+9x^2+11x-8

    We know  (2x-1) is a factor of  f(x)

    So by using long division, divide  f(x) by  (2x-1) , to get the quadratic.
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    (Original post by voldejoe)
    I'm assuming you know how to do polynomial division? If you do, divide your cubic by 2x-1
    is that (2x-1)(ax^2+bx+c)? and do that = stuff between?
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    (Original post by raheem94)
     f(x) = 2x^3+9x^2+11x-8

    We know  (2x-1) is a factor of  f(x)

    So by using long division, divide  f(x) by  (2x-1) , to get the quadratic.
    I don't understand it this time. How do you do long division with quadratics?


    -oops i just double posted X_X sorry -
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    (Original post by Stickyelmo)
    is that (2x-1)(ax^2+bx+c)? and do that = stuff between?
    Do you know algebraic long division?
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    (Original post by Stickyelmo)
    I don't understand it this time. How do you do long division with quadratics?


    -oops i just double posted X_X sorry -
     (2x-1) \sqrt{2x^3+9x^2+11x-8 }

    Do you know how to do this?
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    (Original post by raheem94)
     (2x-1) \sqrt{2x^3+9x^2+11x-8 }

    Do you know how to do this?
    no I know how to do this though:
    2x^3+9x^2+11x-8=(2x-1)(ax^2+bx+c) if that is right. not sure though. Is your way easier?
    try to teach me your way, might be easier than what my teacher taught me i've learnt so much on tsr than in class X_X though i concentrate hard
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    (Original post by Stickyelmo)
    no I know how to do this though:
    2x^3+9x^2+11x-8=(2x-1)(ax^2+bx+c) if that is right. not sure though. Is your way easier?
    Both methods can be used.

    So if you know the way, then what is the problem?
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    (Original post by raheem94)
    Both methods can be used.

    So if you know the way, then what is the problem?
    most of the time in maths I know all the methods but Im doing like a billion practice papers so I know when to use the methods. Most of the time I'm on the verge of the answer but at a mind block
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    (Original post by Stickyelmo)
    no I know how to do this though:
    2x^3+9x^2+11x-8=(2x-1)(ax^2+bx+c) if that is right. not sure though. Is your way easier?
    try to teach me your way, might be easier than what my teacher taught me i've learnt so much on tsr than in class X_X though i concentrate hard
    See this video to learn my method.
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    (Original post by Stickyelmo)
    most of the time in maths I know all the methods but Im doing like a billion practice papers so I know when to use the methods. Most of the time I'm on the verge of the answer but at a mind block
    You should get  2x^3+9x^2+11x-8  = (2x-1)(x^2+5x+8)

    Try to get it.
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    (Original post by raheem94)
    You should get  2x^3+9x^2+11x-8  = (2x-1)(x^2+5x+8)

    Try to get it.
    yeah I finally got it :P didn't see that one initially but I should do next time!

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