Hey there Sign in to join this conversationNew here? Join for free

Factorising to form linear factor and quadratic

Announcements Posted on
Study Help needs new mods! 14-04-2014
Post on TSR and win a prize! Find out more... 10-04-2014
    • Thread Starter
    • 2 followers
    Offline

    ReputationRep:
    How do i factorise a cubic? the equation is:

    2x^3+9x^2+11x-8

    i am giving (2x-1) is a factor.

    so far I have (2x-1)(x^2+5x.....)
    any tips on a how to in an exam would be helpful too

    Thanks in advance
    • 0 followers
    Offline

    ReputationRep:
    Yo, by far my favourite method for polynomials of small degree is something which I've forgotten the name of! Comparison of factors or something?

    Anyway, we write a general quadratic and note (2x-1)(ax^2+bx+c), then we note that the only way to get an x^3 is by the 2x * ax^2, so 2ax^3 = 2x^3 implies a = 1, then we note that the only way to get a constant is -1*c, so -c = -8, c= 8, this leaves the trickier b, which is like, 2x*c - bx right? Soooo
    x(16-b) = 11x, aha! b is 5, just as you suspected.

    This idea can be extended to polynomials of n degree, make a general polynomial of n-1 degree, but with big n this can get messy.

    Edit: Seriously, what is that called, it's killing me.
    • 0 followers
    Offline

    ReputationRep:
    I'm assuming you know how to do polynomial division? If you do, divide your cubic by 2x-1
    • 46 followers
    Offline

    ReputationRep:
    (Original post by Stickyelmo)
    How do i factorise a cubic? the equation is:

    2x^3+9x^2+11x-8

    i am giving (2x-1) is a factor.

    so far I have (2x-1)(x^2+5x.....)
    any tips on a how to in an exam would be helpful too

    Thanks in advance
     f(x) = 2x^3+9x^2+11x-8

    We know  (2x-1) is a factor of  f(x)

    So by using long division, divide  f(x) by  (2x-1) , to get the quadratic.
    • Thread Starter
    • 2 followers
    Offline

    ReputationRep:
    (Original post by voldejoe)
    I'm assuming you know how to do polynomial division? If you do, divide your cubic by 2x-1
    is that (2x-1)(ax^2+bx+c)? and do that = stuff between?
    • Thread Starter
    • 2 followers
    Offline

    ReputationRep:
    (Original post by raheem94)
     f(x) = 2x^3+9x^2+11x-8

    We know  (2x-1) is a factor of  f(x)

    So by using long division, divide  f(x) by  (2x-1) , to get the quadratic.
    I don't understand it this time. How do you do long division with quadratics?


    -oops i just double posted X_X sorry -
    • 46 followers
    Offline

    ReputationRep:
    (Original post by Stickyelmo)
    is that (2x-1)(ax^2+bx+c)? and do that = stuff between?
    Do you know algebraic long division?
    • 46 followers
    Offline

    ReputationRep:
    (Original post by Stickyelmo)
    I don't understand it this time. How do you do long division with quadratics?


    -oops i just double posted X_X sorry -
     (2x-1) \sqrt{2x^3+9x^2+11x-8 }

    Do you know how to do this?
    • Thread Starter
    • 2 followers
    Offline

    ReputationRep:
    (Original post by raheem94)
     (2x-1) \sqrt{2x^3+9x^2+11x-8 }

    Do you know how to do this?
    no I know how to do this though:
    2x^3+9x^2+11x-8=(2x-1)(ax^2+bx+c) if that is right. not sure though. Is your way easier?
    try to teach me your way, might be easier than what my teacher taught me i've learnt so much on tsr than in class X_X though i concentrate hard
    • 46 followers
    Offline

    ReputationRep:
    (Original post by Stickyelmo)
    no I know how to do this though:
    2x^3+9x^2+11x-8=(2x-1)(ax^2+bx+c) if that is right. not sure though. Is your way easier?
    Both methods can be used.

    So if you know the way, then what is the problem?
    • Thread Starter
    • 2 followers
    Offline

    ReputationRep:
    (Original post by raheem94)
    Both methods can be used.

    So if you know the way, then what is the problem?
    most of the time in maths I know all the methods but Im doing like a billion practice papers so I know when to use the methods. Most of the time I'm on the verge of the answer but at a mind block
    • 46 followers
    Offline

    ReputationRep:
    (Original post by Stickyelmo)
    no I know how to do this though:
    2x^3+9x^2+11x-8=(2x-1)(ax^2+bx+c) if that is right. not sure though. Is your way easier?
    try to teach me your way, might be easier than what my teacher taught me i've learnt so much on tsr than in class X_X though i concentrate hard
    See this video to learn my method.
    • 46 followers
    Offline

    ReputationRep:
    (Original post by Stickyelmo)
    most of the time in maths I know all the methods but Im doing like a billion practice papers so I know when to use the methods. Most of the time I'm on the verge of the answer but at a mind block
    You should get  2x^3+9x^2+11x-8  = (2x-1)(x^2+5x+8)

    Try to get it.
    • Thread Starter
    • 2 followers
    Offline

    ReputationRep:
    (Original post by raheem94)
    You should get  2x^3+9x^2+11x-8  = (2x-1)(x^2+5x+8)

    Try to get it.
    yeah I finally got it :P didn't see that one initially but I should do next time!

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?

    this is what you'll be called on TSR

  2. this can't be left blank
    this email is already registered. Forgotten your password?

    never shared and never spammed

  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. By completing the slider below you agree to The Student Room's terms & conditions and site rules

  2. Slide the button to the right to create your account

    Slide to join now Processing…

    You don't slide that way? No problem.

Updated: May 10, 2012
Article updates
Reputation gems:
You get these gems as you gain rep from other members for making good contributions and giving helpful advice.