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Math Help - Rational Root problem

  1. #1
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    Rational Root problem

    How is this problem done, and what answers would you get for it?
    List all possible rational roots for the function


    Give your list in increasing order. Beside each possible rational root, type "yes" if it is a root and "no" if it is not a root. Leave any unnecessary answer blanks empty.

    Possible rational root: Is it a root? .
    Possible rational root: Is it a root? .
    Possible rational root: Is it a root? .
    Possible rational root: Is it a root? .
    Possible rational root: Is it a root? .
    Possible rational root: Is it a root? .
    Possible rational root: Is it a root? .
    Possible rational root: Is it a root? .
    Possible rational root: Is it a root? .
    Possible rational root: Is it a root? .
    Possible rational root: Is it a root? .
    Possible rational root: Is it a root? .
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    You do this

    Using the rational root theorem which states that in a function where p is the constant and q is the leading coefficent that after you show ALL the factors of p and q that the only possible rational real roots are all the factors of p over all the factors of q make sense? so in this case you find the factors of p(i.e. 57) which lucky is is a prime so its factors are \pm57, \pm1 and the factors of q which once again is aprime so the factors are \pm3, \pm1 now take all the p's over q's
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    Using the rational root theorem which states that in a function where p is the constant and q is the leading coefficent that after you show ALL the factors of p and q that the only possible rational real roots are all the factors of p over all the factors of q make sense?
    yeah my friend and I tried that and we got imaginary numbers/answers. That why I am also curious what other people got for the answer
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    What do you mean

    of course there is always the possibility of imaginary roots...but they aren't part of the problem?....wait using Descartes rule of signs there are 4 solutions and they are all real? you must be doing something wrong
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  5. #5
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    Quote Originally Posted by Mathstud28 View Post
    of course there is always the possibility of imaginary roots...but they aren't part of the problem?
    thats what me and my friend also got, sorry math is really not my best subject
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Girlaaaaaaaa View Post
    yeah my friend and I tried that and we got imaginary numbers/answers. That why I am also curious what other people got for the answer
    the problem did not ask you to find the roots, there are no real roots. you were asked to list the POSSIBLE rational roots. Mathstud gave you the correct method. Use the rational roots theorem.

    so find all the possible roots, and then answer "no" to the "is it a root?" question for every one
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    But

    you are right this is wrong because all the answers are imaginary...so the question is a trick...you just want the possible roots not the actual ones!
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  8. #8
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    Quote Originally Posted by Mathstud28 View Post
    you are right this is wrong because all the answers are imaginary...so the question is a trick...you just want the possible roots not the actual ones!
    would it be then? At least one of the answers above is NOT correct.


    (1 pt) List all possible rational roots for the function


    Give your list in increasing order. Beside each possible rational root, type "yes" if it is a root and "no" if it is not a root. Leave any unnecessary answer blanks empty.
    Those answer were said not to be correct
    Possible rational root:-57 Is it a root? .no
    Possible rational root: -19 Is it a root? . no
    Possible rational root:-1 Is it a root? . no
    Possible rational root:-1/3 Is it a root? .no
    Possible rational root: 1/3 Is it a root? . no
    Possible rational root: 1 Is it a root? . no
    Possible rational root: 19 Is it a root? . no
    Possible rational root: 57 Is it a root? . no
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    All you have to do

    is take all your ps and qs and make them the \frac{p}{q}s...that is your answer...none of them are real roots just list the \frac{p}{q}s
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  10. #10
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    Quote Originally Posted by Mathstud28 View Post
    is take all your ps and qs and make them the \frac{p}{q}s...that is your answer...none of them are real roots just list the \frac{p}{q}s
    thats what i did, it did not work I just tried it
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  11. #11
    MHF Contributor Mathstud28's Avatar
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    What

    what do you mean it didnt work...the numbers didnt divide...you do understand that even if a number is a \frac{p}{q} it doesnt neccasarily have to be a root...all those numbers are possible rational roots...but none of them work in this case
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