Results 1 to 8 of 8

Math Help - Need explanation for basic factoring...

  1. #1
    Newbie
    Joined
    Jul 2010
    Posts
    5

    Need explanation for basic factoring...

    Hi guys,

    I've just started a discrete mathematics course at university, and one of the things that is assumed knowledge is being able to factor expressions. The notes then go on to give you some exercises for you to test yourself with. There are accompanied answers, but they don't show how they worked them out. I've got some simple ones solved, but not the ones that are slightly more complicated

    I haven't done anything like this for years and years, so I've forgotten it. I'll show you the example questions so hopefully someone can use them as a reference in explaining them to me.

    Example:

    Let f(x)= $3x^{2} +2 $, <br />
g(y) =  $2y -1$ and <br />
h(z) = $-z^{2} +3 $, express the following as polynomials:
    a) f(y)
    b) g(x)
    c) h(n)
    d) f(m+1)
    e) g(n-2)
    f) h(3k + 1)

    The answers for the first three were f(x)= $3y^{2} +2 $, <br />
g(y) =  $2x -1$ and <br />
h(z) = $-n^{2} +3 $ (really obvious actually, not really doing anything)

    But the other three, I got halfway through but I couldn't find out how the final answer was arrived at. The final answers were given as these:
    d)  f(m+1)= $3m^{2} + 6m +5 $
    e) g(n-2) =  $2n -5$
    f)  h(3k + 1) = $ -9k^{2} - 6k + 2 $

    If I'm being at all vague (which I'm sure I am) please let me know and I'll try to reword the problem I'm having.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Replace x with m+1 therefore,

    f(m+1)= 3(m+1)^2+2= 3(m^2+2m+1)+2= 3m^2+6m+3+2= 3m^2+6m+5

    do you follow?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by Hiram View Post
    Hi guys,

    I've just started a discrete mathematics course at university, and one of the things that is assumed knowledge is being able to factor expressions. The notes then go on to give you some exercises for you to test yourself with. There are accompanied answers, but they don't show how they worked them out. I've got some simple ones solved, but not the ones that are slightly more complicated

    I haven't done anything like this for years and years, so I've forgotten it. I'll show you the example questions so hopefully someone can use them as a reference in explaining them to me.

    Example:

    Let f(x)= $3x^{2} +2 $, <br />
g(y) =  $2y -1$ and <br />
h(z) = $-z^{2} +3 $, express the following as polynomials:
    a) f(y)
    b) g(x)
    c) h(n)
    d) f(m+1)
    e) g(n-2)
    f) h(3k + 1)

    The answers for the first three were f(x)= $3y^{2} +2 $, <br />
g(y) =  $2x -1$ and <br />
h(z) = $-n^{2} +3 $ (really obvious actually, not really doing anything)

    But the other three, I got halfway through but I couldn't find out how the final answer was arrived at. The final answers were given as these:
    d)  f(m+1)= $3m^{2} + 6m +5 $
    e) g(n-2) =  $2n -5$
    f)  h(3k + 1) = $ -9k^{2} - 6k + 2 $

    If I'm being at all vague (which I'm sure I am) please let me know and I'll try to reword the problem I'm having.
    f(y)=3y^2+2

    g(x)=2x-1

    h(n)=-n^2+3

    f(m+1)=3(m+1)^2+2=3(m+1)(m+1)+2=3[m(m+1)+1(m+1)]+2

    g(n-2)=2(n-2)-1

    h(3k+1)=-(3k+1)^2+3=-(3k+1)(3k+1)+3 etc
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jul 2010
    Posts
    5
    Not really, where did the 2m in the second iteration come from?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    (m+1)^2=(m+1)(m+1)=m(m+1)+1(m+1)=m^2+m+m+1=m^2+2m+  1

    Which is based on 5(5)=(3+2)(3+2)=3(3+2)+2(3+2)=3^2+3(2)+2(3)+2(2) and so on...
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jul 2010
    Posts
    5
    I guess what I'm getting at is, I don't know how to factorise. Thanks anyway guys, I'll find someplace else.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Hiram View Post
    I guess what I'm getting at is, I don't know how to factorise. Thanks anyway guys, I'll find someplace else.
    These questions are not about factorisation, but about substituting one expresion (infact a linear expresion) for the variable in another (a polynomial in these cases).
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member eumyang's Avatar
    Joined
    Jan 2010
    Posts
    278
    Thanks
    1
    OP: be careful with the terminology. It's not factoring, it's just multiplying. Factoring is going backwards, in a sense, where you are given a product and you need to find two (or more numbers) multiplied together to get to that product. So doing something like this:
    (x + 3)(x + 2) = x^2 + 6x + 5
    ...which is what you say you don't know, is multiplying, but doing something like this:
    x^2 - 8x + 15 = (x - 3)(x - 5)
    ... is factoring. See the difference?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Basic factoring
    Posted in the Algebra Forum
    Replies: 8
    Last Post: February 22nd 2011, 06:55 AM
  2. Help with basic explanation
    Posted in the Statistics Forum
    Replies: 5
    Last Post: March 13th 2010, 09:25 PM
  3. Basic Factoring question
    Posted in the Algebra Forum
    Replies: 3
    Last Post: June 19th 2009, 11:21 AM
  4. Basic factoring question
    Posted in the Algebra Forum
    Replies: 3
    Last Post: June 2nd 2008, 07:26 PM
  5. Factoring explanation
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 6th 2007, 06:21 AM

Search Tags


/mathhelpforum @mathhelpforum