Just one sec...
Hey! Sign in to get help with your study questionsNew here? Join for free to post

Factorising

Announcements Posted on
Last chance to win! Our 1984 city-break competition closes on Monday. Post now to enter. 22-07-2016
    Offline

    ReputationRep:
    (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
    • Thread Starter
    Offline

    ReputationRep:
    (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.
    Offline

    ReputationRep:
    (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.
    Offline

    ReputationRep:
    (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))
    • Thread Starter
    Offline

    ReputationRep:
    (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
    Offline

    ReputationRep:
    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)
    • Thread Starter
    Offline

    ReputationRep:
    (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?
    • Thread Starter
    Offline

    ReputationRep:
    (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.
    Offline

    ReputationRep:
    (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
    • Thread Starter
    Offline

    ReputationRep:
    (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)
    Offline

    ReputationRep:
    (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.
    Offline

    ReputationRep:
    (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
    • Thread Starter
    Offline

    ReputationRep:
    (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)
    Offline

    ReputationRep:
    (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)
    • Thread Starter
    Offline

    ReputationRep:
    (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?
    Offline

    ReputationRep:
    (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)
    • Thread Starter
    Offline

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

    ReputationRep:
    (Original post by zed963)
    (x+2Y)(-400b+x+2y)
    I would write it as: (x+2y)(x+2y-400b)
    • Thread Starter
    Offline

    ReputationRep:
    (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
    Offline

    ReputationRep:
    (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

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. Oops, you need to agree to our Ts&Cs to register
  2. Slide to join now Processing…

Updated: April 15, 2012
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Poll
Bourne or Bond?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read here first

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups
Study resources

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22

Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.