Results 1 to 4 of 4
Like Tree1Thanks
  • 1 Post By SlipEternal

Thread: Hey, I've got another exercise ...

  1. #1
    Junior Member
    Joined
    May 2012
    From
    slovenia
    Posts
    69
    Thanks
    4

    Hey, I've got another exercise ...

    We have the polynomial $\displaystyle p(x) = x^3 + 7x^2 + (s + 12)x + s$. This $\displaystyle p(x)$ should not have the complex roots (it needs to have just real roots from real number). Find the values $\displaystyle s$.
    How I find this value $\displaystyle s$?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946

    Re: Hey, I've got another exercise ...

    I would probably be inclined to try to factorise it by grouping...

    $\displaystyle \begin{align*} x^3 + 7x^2 + \left( s + 12 \right) x + s &= x^3 + \left( s + 12 \right) s + 7x^2 + s \\ &= x \left( x^2 + s + 12 \right) + 7 \left( x^2 + \frac{s}{7} \right) \end{align*}$

    so now in order to have a common factor, that means we have to have

    $\displaystyle \begin{align*} s + 12 &= \frac{s}{7} \\ 7s + 84 &= s \\ 6s &= -84 \\ s &= -14 \end{align*}$


    Now just to check:

    $\displaystyle \begin{align*} x^3 +7x^2 + \left( -14 + 12 \right) x - 14 &= x^3 + 7x^2 - 2x - 14 \\ &= x^3 - 2x + 7x^2 - 14 \\ &= x \left( x^2 - 2 \right) + 7 \left( x^2 - 2 \right) \\ &= \left( x^2 - 2 \right) \left( x + 7 \right) \\ &= \left( x - \sqrt{2} \right) \left( x + \sqrt{2} \right) \left( x + 7 \right) \end{align*}$


    So $\displaystyle \begin{align*} s = -14 \end{align*}$ definitely works. I don't know if there are any other possible values you could have for s though.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2010
    Posts
    3,562
    Thanks
    1436

    Re: Hey, I've got another exercise ...

    Let $\displaystyle x = y-\dfrac{7}{3}$. Then:

    $\displaystyle \begin{align*}x^3+7x^2+(s+12)x+s & = \left(y-\dfrac{7}{3}\right)^3+7 \left(y-\dfrac{7}{3}\right)^2+(s+12)\left(y-\dfrac{7}{3}\right)+s \\ & = y^3+\left(s-\dfrac{13}{3}\right)y-\dfrac{36s+70}{27}=0\end{align*}$

    Next, let

    $\displaystyle 3ab = s-\dfrac{13}{3}$
    $\displaystyle a^3-b^3 = \dfrac{36s+70}{27}$

    So, solving for $\displaystyle b^3$ will give a quadratic. From that, I think you can tell what values are going to be complex or not...
    Last edited by SlipEternal; Jan 18th 2015 at 05:40 PM.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    May 2012
    From
    slovenia
    Posts
    69
    Thanks
    4

    Re: Hey, I've got another exercise ...

    Why do you say let $\displaystyle x = y - \dfrac{7}{3}$? Also for $\displaystyle s = 0$ works ...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Set exercise
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Jun 19th 2010, 09:37 AM
  2. Exercise
    Posted in the Statistics Forum
    Replies: 1
    Last Post: Nov 25th 2009, 03:59 AM
  3. Help with an exercise
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Jun 10th 2009, 11:35 PM
  4. can you help me with this exercise?
    Posted in the Business Math Forum
    Replies: 0
    Last Post: Nov 8th 2008, 03:18 AM
  5. can you help me with this exercise?
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Oct 26th 2008, 10:57 AM

Search Tags


/mathhelpforum @mathhelpforum