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    I can factorise or at least I assumed I could untill this question came up:

    Factorise the following function:

    Y = 2 - 3x - 2x^2

    Has me well and truly :confused:

    If anyone could show working out for this question with the answer so I can try and apply it myself hopefully figure it out.
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    well you can start by making y=0

    2 - 3x - 2x^2 = 0

    then you can move everything over to the other side

    2x^2 + 3x - 2 = 0

    do you know what to do from here?
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    (Original post by john1234567)
    well you can start by making y=0

    2 - 3x - 2x^2 = 0

    then you can move everything over to the other side

    2x^2 + 3x - 2 = 0

    do you know what to do from here?

    I thought I did I mean its a similar method I use for completing the square by reversing the equation but after having tried it on this question my answer seems to be wrong
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    Smiley fail.

    Take it all over to the other side if you are really that confused.

    You can factorise it by quadratic formula/completing the square if you are really that confused (solve and then put x - ans into a bracket).

    Factorising just requires, a pencil, rubber and a lot of practice.
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    Answer below

    Spoiler:
    Show
    (2x-1)(x+2)
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    (Original post by john1234567)
    well you can start by making y=0

    2 - 3x - 2x^2 = 0

    then you can move everything over to the other side

    2x^2 + 3x - 2 = 0

    do you know what to do from here?
    If you do this, you have to put a -ve sign after you factorise it because y isn't 0
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    (Original post by Shakaa)
    I thought I did I mean its a similar method I use for completing the square by reversing the equation but after having tried it on this question my answer seems to be wrong

    As aforementioned make it quadratic , then find the routs and factorise :yep:
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    ok i'll show you my method

    a = 2
    b = 3
    c = -2

    start off by finding ac

    ac in this case is -4

    now find two integers m and n such that m + n = ac

    in this case, m = 4 and n = -1

    so rewrite the equation as 2x^2 + 4x - x - 2 = 0

    now imagine you've split the LHS down the middle, so you're left with

    2x^2 + 4x
    and
    - x - 2

    factorise 2x^2 + 4x to get 2x(x + 2)

    the (x + 2) will be the bracket for - x - 2 as well, so write it there

    - x - 2 = s(x + 2)

    find a value of s that will make that equation true; in this case, -1

    so we now have 2x(x + 2) + (-1)(x + 2) = 0

    the (x + 2) is common in both terms, so

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

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    (Original post by john1234567)
    ok i'll show you my method

    a = 2
    b = 3
    c = -2

    start off by finding ac

    ac in this case is -4

    now find two integers m and n such that m + n = ac

    in this case, m = 4 and n = -1

    so rewrite the equation as 2x^2 + 4x - x - 2 = 0

    now imagine you've split the LHS down the middle, so you're left with

    2x^2 + 4x
    and
    - x - 2

    factorise 2x^2 + 4x to get 2x(x + 2)

    the (x + 2) will be the bracket for - x - 2 as well, so write it there

    - x - 2 = s(x + 2)

    find a value of s that will make that equation true; in this case, -1

    so we now have 2x(x + 2) + (-1)(x + 2) = 0

    the (x + 2) is common in both terms, so

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


    Thats got me sorted now thanks
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    (Original post by john1234567)
    well you can start by making y=0
    No, you can't assume y=0
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    (Original post by john1234567)
    ok i'll show you my method

    a = 2
    b = 3
    c = -2

    start off by finding ac

    ac in this case is -4

    now find two integers m and n such that m + n = ac

    in this case, m = 4 and n = -1

    so rewrite the equation as 2x^2 + 4x - x - 2 = 0

    now imagine you've split the LHS down the middle, so you're left with

    2x^2 + 4x
    and
    - x - 2

    factorise 2x^2 + 4x to get 2x(x + 2)

    the (x + 2) will be the bracket for - x - 2 as well, so write it there

    - x - 2 = s(x + 2)

    find a value of s that will make that equation true; in this case, -1

    so we now have 2x(x + 2) + (-1)(x + 2) = 0

    the (x + 2) is common in both terms, so

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

    Expand the brackets and you don't get 2-3x-2x^2 so the factorisation is not correct. The correct factorisation is (x+2)(1-2x)
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    (Original post by Shakaa)
    Thats got me sorted now thanks
    The correct and simpler way to factorise this equations is as it follows:

     0=2-3x-2x^{2}
    1)Move everything to the left and you will get  2x^{2} + 3x - 2 = 0
    2)Solve it as quadratic ->  D= 25 then  x_1,x_2 = \frac{-3\pm5}{4}=(-2),\frac{1}{2}
    These are the routs and you have  (x+2)(x-\frac{1}{2}) = (x+2)(2x-1)

    (Original post by steve2005)
    No, you can't assume y=0
    Yes,we can (sounds like Obama )
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    (Original post by etnies)
    The correct and simpler way to factorise this equations is as it follows:

     0=2-3x-2x^{2}
    1)Move everything to the left and you will get  2x^{2} + 3x - 2 = 0
    2)Solve it as quadratic ->  D= 25 then  x_1,x_2 = \frac{-3\pm5}{4}=(-2),\frac{1}{2}
    These are the routs and you have  (x+2)(x-\frac{1}{2}) = (x+2)(2x-1)



    Yes,we can (sounds like Obama )
    The original question was



    and obviously your solution is incorrect. It's always best to look at the original question rather than subsequent posts.

    Obviously, if you are told that y=0 then the equation can be solved by equating each factor to zero.

    However, the original question was simply factorise.
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    This is my method:

    y = 2 - 3x - 2x^2

    Look for factors of the constant term and the x^2 coefficient that will add up to make 3 I write it in a table:

    1 2 | 1
    -------
    2 1 | 2

    The factors of the constant term on the left and factors of x^2 on the right. Then multiply them diagonally and see which two give the x coefficiant. As you can see you can have 2 x 2 and 1 x 1 which when added together can give 3 or 5.

    so now you know its:

    2 | 1
    -----
    1 | 2

    You write a backet, first with the top two numbers (reading from left to right) and second with bottom two)

    So you get (2 x)(1 2x)

    If you expand the x terms, you get x and 4x. Since we want -3x you make 4x negative and x positive, so you put the signs in the brackets so that you when you expand them, you get -4x + x. In this case it would be (2 + x)(1 - 2x)

    Hope that helped in some way.
 
 
 
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