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Factorising

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  1. Offline

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    (Original post by zed963)
    I am also having trouble with this thing

    (x-y)^2-b(x-y)
    maybe it would help you to "substitute" (x-y) with another letter, say t?

    so you'd have
    t^2 - bt

    and then factorise that out, as you understood how to do above.




    you'd get t(t-b)
    and now put back in your substitution for t... t = (x-y)

    so you'd have (x-y)((x-y)-b)

    the same goes for the (x+5) one:
    (x+5)+3(x+5)^2
    let x+5 = t

    so it becomes
    t+3t^2
    factorise:
    t(1+3t)

    put back in t=(x+5)

    (x+5)(1+3(x+5)) and simplify
  2. Offline

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    (Original post by Math12345)
    1. Write out the quadratic like this: (x-y)(x-y)-b(x-y)
    2. Common factor? (...)
    3. Then you collect the other terms in the other bracket (...)
    (x-y) But you see thats the thing I am not able to spot that immeadiately is there any resource that can teach you this thing.
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    (Original post by zed963)
    (x-y) But you see thats the thing I am not able to spot that immeadiately is there any resource that can teach you this thing.
    That's why you should write (x-y)^2 as (x-y)(x-y) - It's much easier to spot the common factor then.
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    (Original post by zed963)
    Now thats the bit where I get confused.
    So

    y+3y^2 = y(1+3y)

    m+3m^2 = m(1+3m)

    ab+3(ab)^2 = ab(1+3ab)

    (Zed)+3(Zed)^2 = Zed(1+3Zed)

    (anything)+3(anything)^2 = (anything)(1+3(anything))

    (x+5)+3(x+5)^2 = (x+5)(1+3(x+5))
  5. Offline

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    (Original post by TenOfThem)
    So

    y+3y^2 = y(1+3y)

    m+3m^2 = m(1+3m)

    ab+3(ab)^2 = ab(1+3ab)

    (Zed)+3(Zed)^2 = Zed(1+3Zed)

    (anything)+3(anything)^2 = (anything)(1+3(anything))

    (x+5)+3(x+5)^2 = (x+5)(1+3(x+5))
    So if we were to expand this whole equation what would it become: (x+5)+3(x+5)^2
  6. Offline

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    If you expand (x+5) + 3(x+5)^2 = x+5 + 3x^2+30x+75 = 3x^2 + 31x + 80

    If you meant how do we fully factorise

    (x+5) + 3(x+5)^2 = (x+5)(1+3(x+5)) = (x+5)(1+3x+15) = (x+5)(3x+16)
  7. Offline

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    (Original post by Math12345)
    That's why you should write (x-y)^2 as (x-y)(x-y) - It's much easier to spot the common factor then.
    So i f i have one common factor then what do i do after?
  8. Offline

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    (Original post by TenOfThem)
    If you expand (x+5) + 3(x+5)^2 = x+5 + 3x^2+30x+75 = 3x^2 + 31x + 80

    If you meant how do we fully factorise

    (x+5) + 3(x+5)^2 = (x+5)(1+3(x+5)) = (x+5)(1+3x+15) = (x+5)(3x+16)
    I get the expand bit but I've been taught to find the factors of 240 that make 31 and then split it in half and factorise, thats why i'm not familiar with the common factor thing.
  9. Offline

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    (Original post by zed963)
    I get the expand bit but I've been taught to find the factors of 240 that make 31 and then split it in half and factorise, thats why i'm not familiar with the common factor thing.
    Oh ... you mean to factorise the

    3x^2 + 31x + 80

    version

    well that is fine ... why not just use that
  10. Offline

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    (Original post by TenOfThem)
    Oh ... you mean to factorise the

    3x^2 + 31x + 80

    version

    well that is fine ... why not just use that
    But I don't think that works for (x-y)^2 -b(x-y)
  11. Offline

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    (Original post by zed963)
    I get the expand bit but I've been taught to find the factors of 240 that make 31 and then split it in half and factorise, thats why i'm not familiar with the common factor thing.
    If they give you the quadratic like that then you should do that, but if they give you a common factor then it is much easier to use that.
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    (x-y)^2 - b(x-y) = (x-y)(x-y-b)

    is clearly the easiest way

    I am not really sure what you are struggling with in the above

    Common factor simply means the same thing in both parts and it is clear that there is an (x-y) in both bits
  13. Offline

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    (Original post by TenOfThem)
    (x-y)^2 - b(x-y) = (x-y)(x-y-b)

    is clearly the easiest way

    I am not really sure what you are struggling with in the above

    Common factor simply means the same thing in both parts and it is clear that there is an (x-y) in both bits
    Yes I get the common factor bit but not how you get (x-y-b)
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    (x-y)(x-y)-b(x-y)
    Common factor: (x-y)

    Cross it out above, so you are left with.
    (x-y)-b

    Put this into one bracket:
    (x-y-b)

    Therefore: (x-y)(x-y-b)
  15. Offline

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    (Original post by Math12345)
    (x-y)(x-y)-b(x-y)
    Common factor: (x-y)

    Cross it out above, so you are left with.
    (x-y)-b

    Put this into one bracket:
    (x-y-b)

    Therefore: (x-y)(x-y-b)
    Oh right thanks, Can you give me a similar expression so that I can factorise it?
  16. Offline

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    (Original post by zed963)
    Oh right thanks, Can you give me a similar expression so that I can factorise it?
    (x+2y)^2-400b(x+2y)
  17. Offline

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    (Original post by Math12345)
    (x+2y)^2-400b(x+2y)
    (x+2Y)(-400bx+2y)
  18. Offline

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    (Original post by zed963)
    (x+2Y)(-400b+x+2y)
    I would write it as: (x+2y)(x+2y-400b)
  19. Offline

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    (Original post by Math12345)
    I would write it as: (x+2y)(x+2y-400b)
    So the above is right,

    So if i had this expression : 6y(x+3y)+9(x+3y)^2

    Would that be (x+3y) and something which I can't figure out HELP
  20. Offline

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    (Original post by zed963)
    So the above is right,

    So if i had this expression : 6y(x+3y)+9(x+3y)^2

    Would that be (x+3y) and something which I can't figure out HELP
    Yes

    What you have there is

    6y(x+3y) + 9(x+3y)(x+3y)

    Take out the red and what is left ... that is your other bracket

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