SOLUTION: 10x^(5)+39x^(4)-24x^(3)-238x^(2)-216x+24=0 find the solution set using rational zero theorem and descarters's rule of signs

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Question 633042: 10x^(5)+39x^(4)-24x^(3)-238x^(2)-216x+24=0
find the solution set using rational zero theorem and descarters's rule of signs

Answer by jsmallt9(3758)   (Show Source): You can put this solution on YOUR website!
(Note: "Root" and "zero" mean pretty much the same thing. So don't get confused when your problem says zero and I keep saying root.)


As a 5th degree polynomial (with real coefficients) there will be 5 roots within the set of Complex numbers.

The possible rational roots of a polynomial are all the possible numbers, positive and negative, that can be formed by a ratio of a factor of the constant term (at the end) over a factor of the leading coefficient (in front of the highest power term).

Your constant term is the 24. Its factors are 1, 2, 3, 4, 6, 8, 12 and 24.

Your leading coefficient is 10. It factors are 1, 2, 5 and 10.

The possible rational roots for your polynomial are all the ratios, positive and negative, that can be formed using a factor of 24 (1, 2, 3, 4, 6, 8, 12 and 24) over a factor of 10 (1, 2, 5 and 10). This list, with duplicates removed and fractions reduced, is:
1, 2, 3, 4, 6, 8, 12, 24, 1/2, 3/2, 1/5, 2/5, 3/5, 4/5, 6/5, 8/5, 12/5, 24/5, 1/10, 3/10 and all the negatives of these.

There are 2 sign changes in the original polynomial (from 39 to -24 and from -216 to 24). From Descartes's rule of signs we know that there will be 2 or zero positive roots.

After changing the signs of the coefficients of the odd-numbered powers of x (-10x^(5)+39x^(4)+24x^(3)-238x^(2)+216x+24), there are 3 sign changes (from -10 to 39, from 24 to -238 and from -238 to 216). From Descartes's rule of signs we know that there will be 3 or 1 negative roots.

There are quite a few possible rational roots to try. But with the information we have from Descartes rule and with some logic and luck we can the roots in a reasonable amount of time. For example, since there may be as many as 3 negative roots but only as many as 2 positive ones, perhaps we will find a negative root faster than a positive one. So let's start by trying the negative rational roots. Checking for a rational root usually easiest with synthetic division (which I hope you've learned since I'm going to use it).
Trying -1:
-1 |  10   39    -24    -238    -216    24
====       -1    -39      63     174    42
      ====================================
      10   38    -63    -174     -42    66

-2 |  10   39    -24    -238    -216    24
====      -20    -38     124     228   -24
      ==================================== 
      10   19    -62    -114      12     0

With a zero remainder, -2 is a root. Not only that, but (x-(-2)) or (x+2) is a factor and the rest of the bottom line is the other factor. "10 19 -62 -114 12" translates into 10x^4+19x^3-62x^2-114x+12. We can use this to our advantage. The remaining roots will be roots of 10x^4+19x^3-62x^2-114x+12 which has a shorter list of possible rational roots (any possible root with a 24 in the numerator is no longer possible). As we continue our search we will be searching for roots of 10x^4+19x^3-62x^2-114x+12. Any roots we tried before which did not work, will not magically work now with 10x^4+19x^3-62x^2-114x+12. But there are multiple roots (or roots of multiplicity n) which means a root can be a root more than once. So I'm going to try -2 again:
-2 |   10   19   -62   -114   12
===        -20     2    120  -12
      ==========================
       10   -1   -60      6    0

Bingo! -2 is a root again. So (x+2) is a factor twice and the remaining factor (from "10 -1 -60 6") is 10x^3-x^2-60x^2+6. 10x^3-x^2-60x^2+6 has many fewer possible roots (any root with an 8 12 or 24 in the numerator are no longer possible). And since we have found two negative roots (the -2 counts twice!) I'm going to switch over to try to find a positive root.
Trying 1:
1  |  10  -1   -60    6
====      10     9  -51
      =================
      10   9   -51  -45
Not a root.
2  |  10  -1   -60    6
====      20    38  -44
      =================
      10  19   -22  -38
Not a root.
3  |  10  -1   -60    6
====      30    87   81
      =================
      10  29    27   87

Not a root. But since the remainder for 3 was positive, 87, and the remainder for 2 was negative. there is a root somewhere between 2 and 3. Unfortunately there are no rational roots of 10x^3-x^2-60x^2+6 between 2 and 3. So that root must be irrational. Next is 4:
4  |  10  -1   -60     6
====      40   156   384
      ==================
      10  39    96   390

Not a root. And since the remainder, 390, is even farther away from zero than 3's remainder I am not hopeful about 6. So I am going to get around to trying some of the fractional rational roots. Looking at the various remainders we've had for 1, 2, 3 and 4, none of them are particularly close to 0. So I am going to try 1/10 since it is the positive possible root that is "farthest" from the others I've tried:
1/10 |  10  -1   -60    6
=====        1     0   -6
       ==================
        10   0   -60    0

A third root! So (x-1/10) is a factor and the remaining factor (from "10 0 -60") is 10x^2-60. Since 10x^2-60 is a quadratic, we can use the Quadratic Formula or other methods to find the remaining two roots. Since there is no "x" term in 10x^2-60, we can just find a square root. Factoring out 10 we get:

From this I hope it is easy to see that (<== this is the one between 2 and 3 that we detected earlier) and are its roots.

So our 5 solutions to

are: -2, -2, 1/10, and

P.S. , in factored form:

or



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