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    Hey this is a question for C1 edexcel.

    How do i factorise this?

    y = 3x + 2x^2 - x^3 = x(3+2x -x^2) << get this bit

    Dont know how to factorise (3+2x -x^2)

    The two numbers which make -3 are 3 and -1. But i dont know how to put them in the brackets because when x is postive i know it looks like (x )(x ) but dont know how it should look for this?

    y=(3-x)(1+x) (this is the answer) why do they have a postive 1 when the numbers are 3 and MiNUS 1 ?

    Thanks
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    (Original post by jayseanfan)
    Hey this is a question for C1 edexcel.

    How do i factorise this?

    y = 3x + 2x^2 - x^3 = x(3+2x -x^2) << get this bit

    Dont know how to factorise (3+2x -x^2)

    The two numbers which make -3 are 3 and -1. But i dont know how to put them in the brackets because when x is postive i know it looks like (x )(x ) but dont know how it should look for this?

    y=(3-x)(1+x) (this is the answer) why do they have a postive 1 when the numbers are 3 and MiNUS 1 ?

    Thanks
    I suggest that you take a minus sign out of the bracket, so that you get -x(x^2-2x-3) and carry on from there.
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    (Original post by tiny hobbit)
    I suggest that you take a minus sign out of the bracket, so that you get -x(x^2-2x-3) and carry on from there.
    thanks but how about if the quesiton was factorise(3+2x -x^2)
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    (Original post by jayseanfan)
    thanks but how about if the quesiton was factorise(3+2x -x^2)
    make it equal to 0 , multiply through by -1 to get x^2 - 2x - 3 which factorises to (x+1) (x-3) ??
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    Reverse it. It now looks like (c+-x)(c+-x) where c is a constant, for example 3, and +- is plus or minus x depending on the signs.

    You're looking for numbers that make 3, not -3, so 3 and 1. You make the x in the first bracket negative so that you get 3x-x which is 2x when you multiply each bracket. You also need to make it negative because you need to get a -x^2.

    Hope this helps lol, you'll get used to factorising quadratics like that eventually.
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    (Original post by jayseanfan)
    thanks but how about if the quesiton was factorise(3+2x -x^2)
    to make -x^2 you have to have (x+a)(-x+b) where a and b are constants

    so from there:

    (-x+3)(x+1)= -x^2-x+3x+3

    this simplifies to: -x^2+2x+3

    they just have the factors in another format in the answer but its exactly the same thing

    Hope this helps happy to help if you have more questions
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    (Original post by mathew551)
    Reverse it. It now looks like (c+-x)(c+-x) where c is a constant, for example 3, and +- is plus or minus x depending on the signs.

    You're looking for numbers that make 3, not -3, so 3 and 1. You make the x in the first bracket negative so that you get 3x-x which is 2x when you multiply each bracket. You also need to make it negative because you need to get a -x^2.

    Hope this helps lol, you'll get used to factorising quadratics like that eventually.
    thanks for the reply, but im still a little confused. Why are we look for numebers whcih make 3 and not -3?

    dont you do a X c which in this case is -1 x 3 = -3 ?
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    You're trying to make C = 3
    as Ax^2 + Bx + C is the general form. Basically 3x1 = 3, but as A must be negative, one of the x's must therefore be negative. As you must then find what adds to make B, which should equal 2, the bracket with 3 in must have a -x in, as when you then expand it, it will be the required quadratic.
    Try expanding the answer and you should see how it works!
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    (Original post by jayseanfan)
    thanks but how about if the quesiton was factorise(3+2x -x^2)
    There's no reason that you can't have a -1 in front of the brackets.

    In spite of all of the other offers, I still suggest that you take out a factor of -1.

    So 3 + 2x - x^2 becomes -1(x^2 -2x-3)

    Now apply your rules to the bracket: factors of -3 are -3 and 1 (to give -2x)

    -1(x-3)(x+1)

    If you don't like the -1 at the front, put it back into one of the brackets e.g. make x-3 into -x+3 or 3-x

    Answer becomes (3-x)(x+1)
 
 
 
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