Results 1 to 6 of 6

Math Help - Factorization Question 2

  1. #1
    Member
    Joined
    Feb 2009
    Posts
    118

    Factorization Question 2

    For the question below, does anyone have a quick and easy method for finding the two numbers that when added together gives 11x, and when multiplied together gives -180? I know the numbers are 20 and -9 but it takes me forever to find them? Is there is quick way?

    \text{Factorize } 12x^2+11x-15
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Furyan's Avatar
    Joined
    Sep 2011
    Posts
    132
    Thanks
    3

    Re: Factorization Question 2

    Hello

    I don't now if this will help, but I do it this way. It doesn't take ages.
    You know you have either:

    1,15 or 3,5

    and your looking for a difference of 11

    Write out:

    (12x )(x )
    (6x )(2x )
    (4x )(3x )

    It doesn't tke long to work out the difference of the possible products and see that 4x5 - 3x3 = 11

    So you have

    (4x - 3)(3x + 5)
    Last edited by Furyan; September 24th 2011 at 04:39 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Factorization Question 2

    If you have a polynomial f(x) and the zero's x_1,x_2,...x_n are known then the polynomial can be factored as:
    f(x)=a(x-x_1)(x-x_2)....(x-x_n) (where a is the coefficient of the highest degree term).
    Offcourse in this case you have to find the zero's first (by using the quadratic formula) ...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,551
    Thanks
    542

    Re: Factorization Question 2

    Hello, sparky!

    For the question below, does anyone have a quick and easy method for
    finding the two numbers whose difference is 11 and whose product 180?
    I know the numbers are 20 and -9, but it takes me forever to find them?
    Is there is quick way?

    \text{Factor: }\:12x^2+11x-15

    I use a primitive approach.
    It takes a bit of time, but it is direct.


    Factor 180 into pairs of factors, and note the difference.

    . . \begin{array}{cc}\text{Factors} & \text{Diff.} \\ \hline 1,180 & 179 \\ 2,90 & 88 \\ 3,60 & 57 \\ 4,45 & 41 \\ 5,36 & 31 \\ 6,30 & 24 \\ 7,? & - \\ 8,? & - \\ 9,20 & 11 \\ 10,18 & 8 \\ 11,? & - \\ 12,15 & 3 \\ 13,? & - \\ \hline \end{array} \begin{array}{c} \\ \\ \\ \\ \\ \leftarrow\text{ Here!} \end{array}


    How do we find these pairs?
    Divide 180 by 1, 2, 3, . . .
    Some of the divisions do not "come out even", of course.

    When can we stop?
    We can stop at the integer part of \sqrt{180} \,\rightarrow\, 13


    We see that the pair (9,20) has a difference of 11.
    (Of course, we can stop listing the moment we find it.)

    We have: . 12x^2 \quad 9x\quad 20x - 15

    We want the middle term to be +11x, so we will use -9x and +20x.

    So we have: . 12x^2 - 9x + 20x - 15

    Factor "by grouping": . 3x(4x-3) + 5(4x-3) \;=\;(4x-3)(3x+5)

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Feb 2009
    Posts
    118

    Re: Factorization Question 2

    Quote Originally Posted by Soroban View Post
    Hello, sparky!


    I use a primitive approach.
    It takes a bit of time, but it is direct.


    Factor 180 into pairs of factors, and note the difference.

    . . \begin{array}{cc}\text{Factors} & \text{Diff.} \\ \hline 1,180 & 179 \\ 2,90 & 88 \\ 3,60 & 57 \\ 4,45 & 41 \\ 5,36 & 31 \\ 6,30 & 24 \\ 7,? & - \\ 8,? & - \\ 9,20 & 11 \\ 10,18 & 8 \\ 11,? & - \\ 12,15 & 3 \\ 13,? & - \\ \hline \end{array} \begin{array}{c} \\ \\ \\ \\ \\ \leftarrow\text{ Here!} \end{array}


    How do we find these pairs?
    Divide 180 by 1, 2, 3, . . .
    Some of the divisions do not "come out even", of course.

    When can we stop?
    We can stop at the integer part of \sqrt{180} \,\rightarrow\, 13


    We see that the pair (9,20) has a difference of 11.
    (Of course, we can stop listing the moment we find it.)

    We have: . 12x^2 \quad 9x\quad 20x - 15

    We want the middle term to be +11x, so we will use -9x and +20x.

    So we have: . 12x^2 - 9x + 20x - 15

    Factor "by grouping": . 3x(4x-3) + 5(4x-3) \;=\;(4x-3)(3x+5)

    Wow, I love this! Why didn't I think of this? This was so obvious! Thanks
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,155
    Thanks
    595

    Re: Factorization Question 2

    Quote Originally Posted by Soroban View Post
    Hello, sparky!


    I use a primitive approach.
    It takes a bit of time, but it is direct.


    Factor 180 into pairs of factors, and note the difference.

    . . \begin{array}{cc}\text{Factors} & \text{Diff.} \\ \hline 1,180 & 179 \\ 2,90 & 88 \\ 3,60 & 57 \\ 4,45 & 41 \\ 5,36 & 31 \\ 6,30 & 24 \\ 7,? & - \\ 8,? & - \\ 9,20 & 11 \\ 10,18 & 8 \\ 11,? & - \\ 12,15 & 3 \\ 13,? & - \\ \hline \end{array} \begin{array}{c} \\ \\ \\ \\ \\ \leftarrow\text{ Here!} \end{array}


    How do we find these pairs?
    Divide 180 by 1, 2, 3, . . .
    Some of the divisions do not "come out even", of course.

    When can we stop?
    We can stop at the integer part of \sqrt{180} \,\rightarrow\, 13


    We see that the pair (9,20) has a difference of 11.
    (Of course, we can stop listing the moment we find it.)

    We have: . 12x^2 \quad 9x\quad 20x - 15

    We want the middle term to be +11x, so we will use -9x and +20x.

    So we have: . 12x^2 - 9x + 20x - 15

    Factor "by grouping": . 3x(4x-3) + 5(4x-3) \;=\;(4x-3)(3x+5)

    ^this. is beautiful.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: December 7th 2011, 04:23 PM
  2. [SOLVED] Factorization Question 1
    Posted in the Algebra Forum
    Replies: 5
    Last Post: September 24th 2011, 11:49 AM
  3. Prime factorization phi question
    Posted in the Number Theory Forum
    Replies: 7
    Last Post: May 6th 2011, 07:26 PM
  4. Prime factorization question
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: November 9th 2009, 11:16 AM
  5. Prime factorization question
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: September 16th 2009, 12:12 PM

/mathhelpforum @mathhelpforum