Results 1 to 8 of 8

Math Help - factorization with binomials

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    12

    factorization with binomials

    Hi everyone,

    I know that the common method is like this :
    x^2 + 5x + 4

    Sum = 4
    product = 5

    so a = 1 and b = 4

    =(x+1)(x+4)

    If I know the product is 36 (9*4) , how does knowing a= -6 b=-6 help me to find the factorized form of (3x-2)(3x-2) ?

    9x^2 -12x +4

    Sum = -12
    product = 36

    a= -6 b=-6

    = 9x^2 -6x-6x+4
    (3x-2)(3x-2)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello masterinex
    Quote Originally Posted by masterinex View Post
    ...
    If I know the product is 36 (9*4) , how does knowing a= -6 b=-6 help me to find the factorized form of (3x-2)(3x-2) ?

    9x^2 -12x +4

    Sum = -12
    product = 36

    a= -6 b=-6

    = 9x^2 -6x-6x+4
    (3x-2)(3x-2)
    Take out the highest common factor from the first pair of terms, and then from the second pair:
    9x^2-6x-6x+4=3x\color{red}(3x-2)\color{black}-2\color{red}(3x-2)
    Then note that the two brackets are the same. So you can now complete the factorisation:
    =(3x-2)\color{red}(3x-2)
    Grandad
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2009
    Posts
    12
    if i know that a = -6 and b =-6 , is there a connection between the greatest common factor 3x and 2 ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Quote Originally Posted by masterinex View Post
    Hi everyone,

    I know that the common method is like this :
    x^2 + 5x + 4

    Sum = 4
    product = 5

    so a = 1 and b = 4

    =(x+1)(x+4)

    If I know the product is 36 (9*4) , how does knowing a= -6 b=-6 help me to find the factorized form of (3x-2)(3x-2) ?

    9x^2 -12x +4

    Sum = -12
    product = 36

    a= -6 b=-6

    = 9x^2 -6x-6x+4
    (3x-2)(3x-2)
    I think you are talking about relating the roots of f(x) by coefficients a,b and c of f(x)=ax^2+bx+c. I believe you are making errors doing this as:

    x_1+x_2= -\frac{b}{a}
    x_1*x_2= \frac{c}{a}

    So if you plug in a, b and c for any particular quadratic equation, you can solve for x_1 and x_2.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2009
    Posts
    12
    I guess the idea is like this :

    9x^2 -12x +4

    Sum = -12
    product = 36

    a and b are just two numbers which added together give -12 and multiplied get 36 . Then the linear term -12x can be rewritten in terms of ax + bx
    a= -6 b=-6

    so 9x^2 -12x +4 becomes
    = 9x^2 -6x-6x+4
    (3x-2)(3x-2)


    for this example the method worked fine . But what about if I have something like this :

    2x^2 + 11x + 5 for which the factored form should be (2x+1)(x+5)

    Then by the method
    product = 10 and sum = 11
    the only two numbers I can think of are 2 and 5 , but the sum wouldnt be 11. Any suggestions ?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Quote Originally Posted by masterinex View Post
    I guess the idea is like this :

    9x^2 -12x +4

    Sum = -12
    product = 36

    a and b are just two numbers which added together give -12 and multiplied get 36 . Then the linear term -12x can be rewritten in terms of ax + bx
    a= -6 b=-6

    so 9x^2 -12x +4 becomes
    = 9x^2 -6x-6x+4
    (3x-2)(3x-2)


    for this example the method worked fine . But what about if I have something like this :

    2x^2 + 11x + 5 for which the factored form should be (2x+1)(x+5)

    Then by the method
    product = 10 and sum = 11
    the only two numbers I can think of are 2 and 5 , but the sum wouldnt be 11. Any suggestions ?
    You are right that when factoring you basically do this method without writing it out, meaning you think of two numbers that add up to something and multiply to something else as well. I don't know of an easy way to use the formulas for sum and product of roots to automatically lead to a factored polynomial, sorry. It might very well exist, but I don't know it.

    Notice that x^2 has a coefficient other than 1. This always complicates the process of factoring. That 2 in front of x^2 is going to multiply with the two terms in the other parentheses but not the term in the same one. So now you aren't trying to find a and b such that a+b is the coefficient of x and ab is the constant term. I know this doesn't really help, but maybe there is something you didn't know buried in this post.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1

    Factorising quadratics

    Hello masterinex
    Quote Originally Posted by masterinex View Post
    I guess the idea is like this :

    9x^2 -12x +4

    Sum = -12
    product = 36

    a and b are just two numbers which added together give -12 and multiplied get 36 . Then the linear term -12x can be rewritten in terms of ax + bx
    a= -6 b=-6

    so 9x^2 -12x +4 becomes
    = 9x^2 -6x-6x+4
    (3x-2)(3x-2)


    for this example the method worked fine . But what about if I have something like this :

    2x^2 + 11x + 5 for which the factored form should be (2x+1)(x+5)

    Then by the method
    product = 10 and sum = 11
    the only two numbers I can think of are 2 and 5 , but the sum wouldnt be 11. Any suggestions ?
    You'll kick yourself - what about 10 and 1?

    Then you can say (using the same method as before):
     2x^2 + 11x  + 5 = 2x^2+10x+x+5
    =2x(x+5)+1(x+5)
    =(2x+1)(x+5)
    Any quadratic that can be factorised will work in the same way, even with numbers with a large number of factors. For example:
    45x^2+126x-80
    Product = 45 \times (-80) = -3600. So we want two factors of 3600 whose difference is 126. A certain amount of trial and error is necessary, but 150 and 24 are the factors required. So:
    45x^2+126x-80
    =45x^2 +150x-24x-80

    =15x(3x+10)-8(3x+10)

    =(15x-8)(3x+10)
    Grandad

    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Oct 2009
    Posts
    12
    owh dang it !
    1 and 10 , that was tricky . So this method really works well, thanks for the great suggestions guys, guess I will drop by to this forum time to time .
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [Factorization] Factorization of polynomials
    Posted in the Algebra Forum
    Replies: 9
    Last Post: April 9th 2010, 12:15 AM
  2. Replies: 2
    Last Post: March 4th 2010, 01:49 AM
  3. binomials
    Posted in the Algebra Forum
    Replies: 2
    Last Post: October 22nd 2008, 01:06 AM
  4. more binomials
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 3rd 2008, 01:32 AM
  5. binomials
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: August 3rd 2005, 06:58 PM

Search Tags


/mathhelpforum @mathhelpforum