document.write( "Question 633042: 10x^(5)+39x^(4)-24x^(3)-238x^(2)-216x+24=0
\n" ); document.write( "find the solution set using rational zero theorem and descarters's rule of signs \n" ); document.write( "
Algebra.Com's Answer #398670 by jsmallt9(3758)\"\" \"About 
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.)

\n" ); document.write( "\"10x%5E%285%29%2B39x%5E%284%29-24x%5E%283%29-238x%5E%282%29-216x%2B24=0\"
\n" ); document.write( "As a 5th degree polynomial (with real coefficients) there will be 5 roots within the set of Complex numbers.

\n" ); document.write( "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).

\n" ); document.write( "Your constant term is the 24. Its factors are 1, 2, 3, 4, 6, 8, 12 and 24.

\n" ); document.write( "Your leading coefficient is 10. It factors are 1, 2, 5 and 10.

\n" ); document.write( "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:
\n" ); document.write( "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.

\n" ); document.write( "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.

\n" ); document.write( "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.

\n" ); document.write( "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).
\n" ); document.write( "Trying -1:
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document.write( "-1 |  10   39    -24    -238    -216    24\r\n" );
document.write( "====       -1    -39      63     174    42\r\n" );
document.write( "      ====================================\r\n" );
document.write( "      10   38    -63    -174     -42    66\r\n" );
document.write( "\r\n" );
document.write( "-2 |  10   39    -24    -238    -216    24\r\n" );
document.write( "====      -20    -38     124     228   -24\r\n" );
document.write( "      ==================================== \r\n" );
document.write( "      10   19    -62    -114      12     0\r\n" );
document.write( "

\n" ); document.write( "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:
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document.write( "-2 |   10   19   -62   -114   12\r\n" );
document.write( "===        -20     2    120  -12\r\n" );
document.write( "      ==========================\r\n" );
document.write( "       10   -1   -60      6    0\r\n" );
document.write( "

\n" ); document.write( "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.
\n" ); document.write( "Trying 1:
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document.write( "1  |  10  -1   -60    6\r\n" );
document.write( "====      10     9  -51\r\n" );
document.write( "      =================\r\n" );
document.write( "      10   9   -51  -45\r\n" );
document.write( "Not a root.\r\n" );
document.write( "2  |  10  -1   -60    6\r\n" );
document.write( "====      20    38  -44\r\n" );
document.write( "      =================\r\n" );
document.write( "      10  19   -22  -38\r\n" );
document.write( "Not a root.\r\n" );
document.write( "3  |  10  -1   -60    6\r\n" );
document.write( "====      30    87   81\r\n" );
document.write( "      =================\r\n" );
document.write( "      10  29    27   87\r\n" );
document.write( "

\n" ); document.write( "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:
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document.write( "4  |  10  -1   -60     6\r\n" );
document.write( "====      40   156   384\r\n" );
document.write( "      ==================\r\n" );
document.write( "      10  39    96   390\r\n" );
document.write( "

\n" ); document.write( "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:
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document.write( "1/10 |  10  -1   -60    6\r\n" );
document.write( "=====        1     0   -6\r\n" );
document.write( "       ==================\r\n" );
document.write( "        10   0   -60    0\r\n" );
document.write( "

\n" ); document.write( "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:
\n" ); document.write( "\"10%28x%5E2-6%29\"
\n" ); document.write( "From this I hope it is easy to see that \"x+=+sqrt%286%29\" (<== this is the one between 2 and 3 that we detected earlier) and \"x+=+-sqrt%286%29\" are its roots.

\n" ); document.write( "So our 5 solutions to
\n" ); document.write( "\"10x%5E%285%29%2B39x%5E%284%29-24x%5E%283%29-238x%5E%282%29-216x%2B24=0\"
\n" ); document.write( "are: -2, -2, 1/10, \"sqrt%286%29\" and \"-sqrt%286%29\"

\n" ); document.write( "P.S. \"10x%5E%285%29%2B39x%5E%284%29-24x%5E%283%29-238x%5E%282%29-216x%2B24\", in factored form:
\n" ); document.write( "\"10%28x%2B2%29%28x%2B2%29%28x-1%2F10%29%28x-sqrt%286%29%29%28x%2Bsqrt%286%29%29\"
\n" ); document.write( "or
\n" ); document.write( "\"%28x%2B2%29%28x%2B2%29%2810x-1%29%28x-sqrt%286%29%29%28x%2Bsqrt%286%29%29\"

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