Results 1 to 7 of 7

Math Help - Polynomial Problem and not a clue how to proceed

  1. #1
    Newbie
    Joined
    Feb 2011
    Posts
    12

    Polynomial Problem and not a clue how to proceed

    I'm on my last problem for my lesson and I'm a little stumped and need a push in the right direction.

    Problem: "One factor of 4x^3+15x^2-31x-30 is x-2. Find the other factors."

    I don't want the answer but if someone could explain to me how they even got x-2 as a factor I should be able to figure the rest out or if someone knows of a lesson that explains this that would be helpful as well.

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

  2. #2
    Super Member TheChaz's Avatar
    Joined
    Nov 2010
    From
    Northwest Arkansas
    Posts
    600
    Thanks
    2
    "How they got" that factor isn't (per se) relevant to how you should proceed, but I'll tell you anyway
    There's a result known as the "rational root theorem" which gives us a finite amount of possible linear factors to check.
    There's also calculators, which can crank out solutions (factors) in no time.

    The big picture is:
    a) you know how to factor/solve quadratics
    b) this is a cubic
    c) dividing a cubic by a (given) linear factor will produce a quadratic, which you know how to factor!

    So we proceed with synthetic division. Set x - 2 = 0 to get x = 2 as the number "in the box"
    [2] 4 15 -31 -30
    ......... 8_ 46_ 30
    ....4... 23.. 15 ...|0|

    Now your quadratic is 4x^2 + 23x + 15 = (4x + 3)(x + 5)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Dec 2009
    From
    1111
    Posts
    872
    Thanks
    3
    Quote Originally Posted by Substince View Post
    I'm on my last problem for my lesson and I'm a little stumped and need a push in the right direction.

    Problem: "One factor of 4x^3+15x^2-31x-30 is x-2. Find the other factors."

    I don't want the answer but if someone could explain to me how they even got x-2 as a factor I should be able to figure the rest out or if someone knows of a lesson that explains this that would be helpful as well.

    Thanks for looking!
    By the Factor theorem. It states that "a polynomial f(x) has a factor (x − k) if and only if f(k) = 0." Therefore given a polynomial you can switch some values and check if the value of f(x) is equal to zero.

    In this problem, f(x)=4x^3+15x^2-31x-30

    f(2)=4.2^3+15.2^2-31.2-30=32+60-62-30=0

    Hence by the Factor theorem, (x-2) is a factor. If you want more details refer, Factor theorem - Wikipedia, the free encyclopedia.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,373
    Thanks
    1315
    Well, I suspect that "they" know that x- 2 is a factor because they created the polynomial by multiplying a quadratic by x- 2! Of course, you could tell that x- 2 is a factor by noting that 4(2)^3+ 15(2)^2- 31(2)- 30= 32+ 60- 62- 30= 92- 92= 0. That tells you that x- 2 evenly divides the polynoial- if it did not we would have P(x)= (x- 2)Q(x)+ r where r is the remainder. Taking x= 2 in that would give P(2)= (0)Q(x)+ r= r. Since P(2)= 0, there is no remainder.

    But the fact that you were told that x- 2 is a factor is all you really needed to know. Now, divide the given polynomial by x- 2 (or use "synthetic division" since the divisor is of the form x- a).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Feb 2011
    Posts
    12
    lol wow do I ever feel stupid right now :P

    Thanks for all the comments guys and clearing things up for me
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,085
    Thanks
    67
    Quote Originally Posted by Substince View Post
    lol wow do I ever feel stupid right now :P
    WHY? You should feel "good" as you know you'll never have that problem again !
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,373
    Thanks
    1315
    But "feeling good" and "feeling stupid" are, fortuately, not mutually exclusive! I learned that a long time ago!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Simple problem but no clue
    Posted in the Geometry Forum
    Replies: 2
    Last Post: January 6th 2010, 02:22 AM
  2. Replies: 2
    Last Post: April 23rd 2009, 11:41 AM
  3. Replies: 3
    Last Post: August 30th 2008, 06:17 AM
  4. I have 4^(x+1) .. how to proceed
    Posted in the Algebra Forum
    Replies: 10
    Last Post: March 24th 2008, 02:21 PM
  5. Calculus Integration problem. NO clue
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 11th 2008, 08:46 PM

Search Tags


/mathhelpforum @mathhelpforum