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    how to find all real solutions to polynomial equations

    6x^6+x^5+29x^4+5X^3+19x^2+4x-4=0

    0  Views: 411 Answers: 2 Posted: 11 years ago

    2 Answers

    The first thing you want to do is find out how many there are.  you can determine this by looking at the sign changes of both f(x) and of f(-x).  in your example there is only going to be one real positive zero because it only has one sign change in f(x).  Now when you look at f(-x) there are five sign changes so there is a possibly five negative real zeros or three or one.  this means that there are possibly no complex zeros or two or three.  anytime that you one or none for pos or neg reals then that is what you have and it will not change that is why the pos real zero doesn't have any other possibilities.  once you have decided the number of possible zeros then it is time to find them.  to do this you will need to know how to do long division of polynomials or synthetic division.  it is a bit of trial and error but you will need to start out with the first coefficient of the equation and the constant at the end these are referred to as P and Q respectively.  6 and -4  you will use all combination of these two numbers and their prime components in the fraction P/Q. P= 1,2,3,6 and Q=+-1,+-2,+-4  thus you will need to divide the equation or expression by +-1/1 or +-1, +-1/2, +-1/4, +-2/1 or +-2 , +-2/2 or +-1, +-2/4 or +-1/2, +-3/1 or +-3, +-3/2, +-3/4, +-6/1 or +-6, +-6/2 or +-3, +-6/4 or +-3/2.  notice that some of them repeat this means that one of these choices could have a multiplicity higher than one so you will need to try to divide them in more than once. for instance


    x^3+3x^2+3x +1 if you divide this once using (x+1) you will get x^2+2x+1  which can then be divided by (x+1) again giving you (x+1) thus your final answer is (x+1)^3 which means you have one real neg zero however it has a multiplicity of 3 and that is why the beginning of this process are only possible quantities of real and complex zeros.  just remember to try all of the P/Q in your division so that you are sure to break the expression done completely so you can get all of your zeros.   Hope this wasn't to confusing sorry i forgot about the fact that it was a -4 so that doubles the possibilities 

    The first thing you want to do is find out how many there are.  you can determine this by looking at the sign changes of both f(x) and of f(-x).  in your example there is only going to be one real positive zero because it only has one sign change in f(x).  Now when you look at f(-x) there are five sign changes so there is a possibly five negative real zeros or three or one.  this means that there are possibly no complex zeros or two or three.  anytime that you one or none for pos or neg reals then that is what you have and it will not change that is why the pos real zero doesn't have any other possibilities.  once you have decided the number of possible zeros then it is time to find them.  to do this you will need to know how to do long division of polynomials or synthetic division.  it is a bit of trial and error but you will need to start out with the first coefficient of the equation and the constant at the end these are referred to as P and Q respectively.  6 and 4  you will use all combination of these two numbers and their prime components in the fraction P/Q. P= 1,2,3,6 and Q=1,2,4  thus you will need to divide the equation or expression by 1/1 or 1, 1/2, 1/4, 2/1 or 2 , 2/2 or 1, 2/4 or 1/2, 3/1 or 3, 3/2, 3/4, 6/1 or 6, 6/2 or 3, 6/4 or 3/2.  notice that some of them repeat this means that one of these choices could have a multiplicity higher than one so you will need to try to divide them in more than once. for instance


    x^3+3x^2+3x +1 if you divide this once using (x+1) you will get x^2+2x+1  which can then be divided by (x+1) again giving you (x+1) thus your final answer is (x+1)^3 which means you have one real neg zero however it has a multiplicity of 3 and that is why the beginning of this process are only possible quantities of real and complex zeros.  just remember to try all of the P/Q in your division so that you are sure to break the expression done completely so you can get all of your zeros.   Hope this wasn't to confusing



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