Results 1 to 9 of 9

Math Help - More Complex Factorisation

  1. #1
    Member
    Joined
    Jan 2009
    Posts
    98

    More Complex Factorisation

    How can I factorise the following? I am new to more complex forms of factorisation and am teaching myself at home.

    1. 4p^5q^3 + 6p^2q^4
    2. 3(x+6)^2 - 9(x-6)
    3. 12a^4 - 3
    4. 36p^3 - 25pq^2
    5. 2x^2 + x -21
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    honolulu
    Posts
    580
    Quote Originally Posted by JadeKiara View Post
    How can I factorise the following? I am new to more complex forms of factorisation and am teaching myself at home.

    1. 4p^5q^3 + 6p^2q^4
    2. 3(x+6)^2 - 9(x-6)
    3. 12a^4 - 3
    4. 36p^3 - 25pq^2
    5. 2x^2 + x -21
    the first one is
    4p^5q^3 + 6p^2q^4
    2p^2q^3(2p^3 + 3q)

    not sure if the 2nd one is typed ok

    the 3rd one is

    12a^4 - 3
    3 (2 a^2-1) (2 a^2+1)

    do you see what is happening???
    Last edited by bigwave; December 3rd 2009 at 10:32 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2009
    Posts
    98

    Thank you very much, bigwave!

    By the way, the 2nd one is in its original form from the textbook.




    Ah! This is definitely an aid in my understanding. But I just have to become accustomed to putting into practice by myself.

    So, correct me if I am wrong:
    (6p^2-5pq)(6p+5q)

    No, it's not right. I am baffled by this one.


    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    honolulu
    Posts
    580

    as for the 4th

     <br />
36p^3 - 25pq^2<br /> <br />

    first factor out what has exponents
    so factor out p

    p\left(36p^2 - 25q^2\right)

    this will factor into a difference of squares

     <br />
p\left(6p - 5q\right)\left(6p + 5q\right)<br />

    has the fog cleared
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Jun 2009
    From
    Africa
    Posts
    641

    Smile

    the last one,
    2x^2 + x -21=(x-3)(2x+7)..
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    honolulu
    Posts
    580

    Talking

    looks good to me
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Jan 2009
    Posts
    98
    Any ideas about number 2?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Jan 2009
    Posts
    98
    Quote Originally Posted by Raoh View Post
    the last one,
    2x^2 + x -21=(x-3)(2x+7)..
    Yes, I did manage to work this one out by myself. All except number 2.

    Thanks for your help!
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    honolulu
    Posts
    580
    Quote Originally Posted by JadeKiara View Post
    Yes, I did manage to work this one out by myself. All except number 2.

    Thanks for your help!
    given:
    3(x+6)^2 - 9(x-6)

    first expand the brackets

    3(x^2 +12x +36) - 9x +54

    3x^2 +36x +108 - 9x + 54

    combine like terms

     <br /> <br />
3x^2 + 27x + 162<br />

    factor out 3

    3(x^2 + 9x +54)

    hope this helps..
    Last edited by bigwave; December 4th 2009 at 03:40 PM. Reason: math correction
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. integrate without complex factorisation
    Posted in the Calculus Forum
    Replies: 7
    Last Post: January 6th 2011, 06:28 AM
  2. complex numbers factorisation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 25th 2009, 06:02 PM
  3. Factorisation
    Posted in the Algebra Forum
    Replies: 4
    Last Post: November 30th 2009, 03:09 AM
  4. Complex Factorisation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 8th 2008, 07:08 AM
  5. factorisation
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 4th 2008, 06:16 PM

Search Tags


/mathhelpforum @mathhelpforum