Results 1 to 5 of 5

Math Help - polynomial functions of Higher Degree

  1. #1
    Newbie
    Joined
    Aug 2010
    Posts
    4

    polynomial functions of Higher Degree

    Hello I am asked the following: "Find the zeros algebraically".

    y=4x^3+4x^2-7x+2

    I know with the four terms I am supposed to group, however Grouping does not work with this problem. How do I factor this to find the zeros. Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    Setting x=-2 You obtain y=0... that means that 4\ x^{3} + 4\ x^{2} - 7\ x + 2 contains x+2 as a factor... etc...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2010
    Posts
    4
    Quote Originally Posted by chisigma View Post
    Setting x=-2 You obtain y=0... that means that 4\ x^{3} + 4\ x^{2} - 7\ x + 2 contains x+2 as a factor... etc...

    Kind regards

    \chi \sigma
    Thank you very much for your reply, however I am wondering how exactly did you find that one of the zeros is -2?

    Thanks
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    Quote Originally Posted by Mack4545 View Post
    Thank you very much for your reply, however I am wondering how exactly did you find that one of the zeros is -2?

    Thanks
    If You insert x=-2 into the expression y(x)= 4\ x^{3} + 4\ x^{2} - 7\ x + 2 You obtain  y(-2) = -32 + 16 +14 +2 =0 ...

    Now You devide y(x)= 4\ x^{3} + 4\ x^{2} - 7\ x + 2 by x+2 and obtain a second degree polynomial whose zeroes can be found in a weel known way...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,429
    Thanks
    1860
    Very useful here is the "rational roots theorem": If m/n is a rational root of a polynomial equation with integer coefficients, then the numerator, m, must evenly divide the constant term and the denominator, n, must evenly divide the leading coefficient.

    Of course, that doesn't guarentee that there are any rational roots but it is always worth trying.

    Here the leading coefficient is 4 so any rational root must have denominator \pm 1, \pm 2, or \pm 4. The constant term is 2 so any rational root must have numerator \pm 1 or \pm 2. That means that the only possible rational roots are \pm 1, \pm \frac{1}{2}, \pm\frac{1}{4}, \pm 2, \pm\frac{2}{2}= \pm 1, and \pm\frac{2}{4}= \pm\frac{1}{2}. Trying each of those shows that only x= -2 really is a root.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Inequality with higher degree
    Posted in the Algebra Forum
    Replies: 8
    Last Post: July 25th 2011, 06:41 PM
  2. Replies: 0
    Last Post: April 26th 2010, 01:50 AM
  3. Replies: 0
    Last Post: March 3rd 2010, 01:25 PM
  4. Replies: 2
    Last Post: March 4th 2009, 12:27 PM
  5. Higher degree equations
    Posted in the Algebra Forum
    Replies: 8
    Last Post: September 23rd 2007, 10:24 AM

Search Tags


/mathhelpforum @mathhelpforum