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Math Help - Finding the Zeros

  1. #1
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    Finding the Zeros

    Find all zeros of the givin polynomial:

    x^4 + x^3 - 6x - 14x - 12
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    Quote Originally Posted by Trentt View Post
    Find all zeros of the givin polynomial:

    x^4 + x^3 - 6x - 14x - 12
    By Rational Roots theorem we check:
    pm 1,pm 2, pm 3, pm 4, pm 6, pm 12

    With luck we find the,
    x=-2 and x=3
    Thus,
    (x+2)(x-3)=x^2-x-6

    Divide,
    Code:
       x^4 + x^3 - 6x^2 - 14x - 12: x^2 - x -6 = x^2+2x+2
    - x^4  +x^3  +6x^2
    ------------------------
                 2x^3 -14 x -12
                -2x^3+2x^2+12x
                ---------------------
                           2x^2 -2x - 12
                           -2x^2+2x-12
                            ---------------
                                             0
    Nice find the zeros of the polynomial,
    x^2+2x+2
    Use quadradic formula to complete this problem.
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  3. #3
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    I have no idea what you just did, once again I'll wait for someone else...thanks anyways..
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  4. #4
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    Re:

    Hi Trent. I worked the problem out for you the answers are x = =2 & x = 3.

    The best way to look at this problem is to graph the function then see where it crosses the x-axis.






    Last edited by qbkr21; March 20th 2007 at 05:58 PM.
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  5. #5
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    My textbook says the solutions are also -1 + i and -1 - i
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Trentt View Post
    My textbook says the solutions are also -1 + i and -1 - i

    there are 4 solutions, -2, 3, -1 + i and -1 - i. you should specify which you want (which evidently are the imaginary solutions). solving the last polynomial which TPH ended with will give you the answers you seek. what exactly didn't you understand about his solution?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Trentt View Post
    I have no idea what you just did, once again I'll wait for someone else...thanks anyways..
    Do you know how to do long division of polynomials Trentt? You'll ned that for this problem, which is what TPH used. An alternate method would be synthetic division.

    Anyway, the rational roots theorem says:

    Given a polynomial a_n*x^n + a_(n-1)*x^(n-1) + ... + a_1*x + a_0 = 0
    the only rational roots are of the form p/q where p divides a_0 and q divides a_n.

    Basically what that means is that you take all the factors of the lone constant and all the factors of the coefficient of the largest power of x and put them in fractions, with the factors of the constant on top, and the factors of the coefficient of x^n on the bottom.

    in this problem, the constant is 12 (who's factors are pm 12, pm 6, pm 4, pm 3, pm 2, pm 1). Note that pm means "plus or minus"

    the coefficient of the highest power of x is 1 (who's factors are pm 1)

    so the zeroes will be (all of these are +/-) 12/1 , 6/1, 4/1, 3/1 , 2/1 and 1/1

    now we take turns plugging each into the equation.

    as TPH says, plugging in -2 and 3 ended up being roots. this means that (x + 2) and (x - 3) are factors of the equation.

    first we divide the equation by x + 2, qbkr21 did this for you with synthetic division, the other method (which i always use--force of habit) is long division.

    anyway, the resulting quotient is x^3 - x^2 - 4x - 6
    so we can rewrite our polynomial as (x + 2)(x^3 - x^2 - 4x - 6)

    now we break down the cubic further by dividing it by (x - 3). the resulting quotient is x^2 + 2x + 2

    so we can write the original polynomial as (x + 2)(x - 3)(x^2 + 2x + 2) = 0

    so now x + 2 = 0 , x - 3 = 0, x^2 + 2x + 2 = 0

    for the quadratic, we use the quadratic formula

    => x = [-2 +/- sqrt(4 - 4(2))]/2
    => x = [-2 +/- sqrt(-4)]/2
    => x = [-2 +/- 2i]/2
    => x = -2/2 +/- 2i/2
    => x = -1 +/- i ..........as you desired

    so there are only 2 real roots, which we can see from the graph qbkr21 posted

    are you still lost?
    Last edited by Jhevon; March 20th 2007 at 06:11 PM.
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  8. #8
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    Yeah, I am. I think they want synthetic division, esp. since I've never seen the rational root theorem or whatever it is.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Trentt View Post
    Yeah, I am. I think they want synthetic division, esp. since I've never seen the rational root theorem or whatever it is.
    ok, well qbkr21 did the synthetic division for you. forget the rational roots theorem then (at least for the time being), i never really learnt it either until i went into an advanced math class. for your purposes just remember to look for the factors of the constant.

    the factors of 12 are +/- 1,2,3,4,6,12
    we begin by plugging these into the original polynomial. why? to figure out which one gives us a zero. why? becuase the question told us to, and finding one makes finding the rest easier.

    so we figure out that -2 and 3 gives zero when we plug them into the equation, so we already found two zeroes. however this is an x^4 graph, so we expect to find four. none of the other numbers we came up with (+/- 1,4,6,12) work. so we know the other zeroes are complex numbers. now to find them.

    we have x = -2 and x = 3 gives zero
    => x + 2 = 0 and x - 3 = 0, so we take x + 2 and x - 3 as factors of the polynomial. that means if we divide by one of them, the remainder is zero and we break down the polynomial to one less power.

    qbkr and TPH already did that and you notice the final result was the polynomial x^2 + 2x + 2 = 0, solving this (which i did earlier) gives you the other 2 zeroes

    Summary:

    1 - Plug in the factors of the lone constant to find what values of x give 0 (note that you don't have to find more than one at a time, find one and run with it)
    2 - if you find some constant k to be a root (a number that gives zero) then x - k is a factor of the polynomial. divide by this factor
    3 - take the quotient and find another facor in the same way, and divide again. repeat this step until you end up with a polynomial
    4 - take all the factors you found and solve for x, these are the zeroes of the polynomial
    Last edited by Jhevon; March 20th 2007 at 06:09 PM. Reason: I left something out in the summary
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    What are you thanking me for qbkr21? you already know how to do the problem

    do you understand now Trentt, i wouldn't want to get thanks for something i couldn't explain well.
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  11. #11
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    I do understand now, thank you
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