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    Edit, I have just realised that putting them either way would work, my bad! I'm not sure if I can delete this thread??? If I can, can anyone tell me how please?

    Hello


    When factorising quadratics, my method is to multiply the first number, by the last number. Then find 2 numbers that multiply to equal that number, and add to make the middle number.
    So this one

    2x^2 + x - 3

    So I would multiply 2 by -3 to get -6.
    Then find two numbers that multiply to together to equal -6, and add together to equal 1 (x).

    So I come up with 3 x - 2 = -6 and 3 + -2 = 1.
    (Another way would be -2 x 3 = 06 and -2 + 3 = 1)

    So I then plug this in and my equation becomes

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

    Then try to factorise to get

    x(2x+3)... and the next bit (-2x - 3) can I guess, only become -1(2x + 3)... Then I use the other method and get

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

    Obviously the second way is the right way, but I feel I will make the same mistake and get the wrong answer. I hope I make sense.

    I suppose what I am asking is, is there a rule or a way of knowing that -2 + 3 was the right way to do it, and not 3 + -2 I originally thought. I suppose how would you know which way to do it, since 3 x -2 = -6 and 3+ -2 = 1, yet -2 x 3 = -6 and -2 + 3 is also 6.

    I probably make no sense but I am hoping someone can give me a nudge on how I would know which way to stick my - and +


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

    It's what multiplies to make the last number and adds up to make the second, not sure why you multiplied the first and last together?
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    (Original post by Tinkletea)
    ...
    The trick to factorisation is spotting it. There is an algorithm that can be used.
    \ x^2 + (A+B)x +AB You form an equation using the values given and solve etc. It's a little more complicated if there is a number in front of the x2 coefficient, but its doable.
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    (Original post by SheldonWannabe)
    2x^2 + x - 3 = (2x+3)(x-1)

    It's what multiplies to make the last number and adds up to make the second, not sure why you multiplied the first and last together?
    My impression is multiplying the first, in this case there is a 2 before the x, and the last, to get 2 x -3 = -6. Then finding 2 numbers that multiply to give -6, and that add to give 1

    Which is;

    -2 x 3 = -6
    -2 + 3 = 1

    Is this incorrect? I did get the same answer as you this way, as I replaced the second number with the above numbers, so it went from

    2x^2 + x - 3

    to

    2x^2 - 2x + 3x -3, then I factorise the first two and get
    2x(x-1) and then factorise the second part and get 3(x-1),

    thus getting (2x+3)(x-1)

    Or, if I reverse -2 and 3 I get

    2x^2 + 3x - 2x - 3, factorise again to get

    x(2x+3) and -1(2x+3), again getting (2x+3)(x-1), when I wrote the post, I didn't think I could factorise -2x - 3, but then realised that -1 was a factor of both, I realised this after and saw that I got the same answer regardless of the way I positioned -2 and 3.

    I'm probably making no sense, but it makes sense to me
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    (Original post by Tinkletea)
    My impression is multiplying the first, in this case there is a 2 before the x, and the last, to get 2 x -3 = -6. Then finding 2 numbers that multiply to give -6, and that add to give 1

    Which is;

    -2 x 3 = -6
    -2 + 3 = 1

    Is this incorrect? I did get the same answer as you this way, as I replaced the second number with the above numbers, so it went from

    2x^2 + x - 3

    to

    2x^2 - 2x + 3x -3, then I factorise the first two and get
    2x(x-1) and then factorise the second part and get 3(x-1),

    thus getting (2x+3)(x-1)

    Or, if I reverse -2 and 3 I get

    2x^2 + 3x - 2x - 3, factorise again to get

    x(2x+3) and -1(2x+3), again getting (2x+3)(x-1), when I wrote the post, I didn't think I could factorise -2x - 3, but then realised that -1 was a factor of both, I realised this after and saw that I got the same answer regardless of the way I positioned -2 and 3.

    I'm probably making no sense, but it makes sense to me
    An interesting way of factorising.
    This method comes in handy in STEP problems and the like, so keep it in mind
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    What is STEP problems, sorry?

    I'm self teaching so I am literally just going through step by step, so don't understand any terms, my motto is getting by, by doing the bare minimum
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    (Original post by Tinkletea)
    ...
    To clarify for you
    The method is fine and you can go either way

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

    2x^2 - 2x + 3x - 3 = 2x(x-1) + 3(x-1) = (2x+3)(x-1)
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    (Original post by Tinkletea)
    What is STEP problems, sorry?

    I'm self teaching so I am literally just going through step by step, so don't understand any terms, my motto is getting by, by doing the bare minimum
    Sixth Term Examination Papers. Used to test for undergraduate entry for top universities in maths. Challenging and rewarding at the same time
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    Ah, I just need a B, I don't like Maths so I definitely won't be going to that level

    Thank you to both of you for your clarification, I will be sure to be more logical before posting a thread next time
 
 
 
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