A valuable fact about the problem's equation: the roots are integers.
x 1 2 + 1 9 x 1 1 − 2 1 8 3 x 1 0 − 3 1 7 1 9 x 9 + 1 5 8 9 4 8 4 x 8 + 1 6 0 9 4 6 2 2 x 7 − 4 5 5 3 8 3 2 3 0 x 6 − 2 9 7 3 0 4 6 8 3 0 x 5 + 4 8 1 2 8 7 5 8 7 6 1 x 4 + 1 7 5 5 8 3 0 0 2 3 0 3 x 3 − 1 4 1 0 3 3 7 8 7 1 5 3 1 x 2 − 1 9 1 6 1 3 8 6 8 4 5 0 7 x + 7 4 2 0 7 3 8 1 3 4 8 1 0 = 0
Note 1: Read the title again!
Note 2: The answer is an integer by value. It was made a floating point number for the answer on a whim.
Note 3: 1 is greater than -2.
Note 4: The absolute value of any of the roots does not exceed 100.
Note 5: The root of the equation: x − 1 = 0 is 1.
This problem's question: What is the largest root by numeric value?
The answer is 31.0.
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1 solution
I think you also have to note that 1 and − 1 are not roots before jumping to knowing the absolute values of all the roots. But that is easily done by inspection.
eg the roots could be − 1 , − 1 , 6 , − 1 1 , 1 3 , − 1 7 , 1 9 , − 2 3 , 2 9 , − 3 1 , 3 5 , − 3 7 adding to − 1 9 .
I guess all you need to say is that the last term is greater than the sum of the absolute values of the others.
The degree of the polynomial is 12. The equation has 12 roots. The constant has 12 prime factors. Therefore, ignoring their signs, those are the roots. Therefore, x=-1 or 1 do not happen in this case.
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No, you need to check 1 and − 1 before you make that conclusion:
Simple case: 2 1 0 = 2 ∗ 3 ∗ 5 ∗ 7 . 4 Prime factors. Below are 2 different polynomials of degree 4 with 4 integer roots and 2 1 0 as the constant:
1) x 4 − 7 x 3 − 1 9 x 2 + 1 6 3 x − 2 1 0 has integer roots: 2 , 3 , − 5 , 7 .
2) x 4 − 7 x 3 − 5 5 x 2 + 2 7 1 x − 2 1 0 has integer roots: 1 , 3 , − 7 , 1 0 .
With 1) you can see that 2 1 0 > 1 6 3 + 1 9 + 7 + 1 = 1 9 0 , so, up to their respective signs, the roots are 2 , 3 , 5 , 7 . The only way for the roots to add up to 7 is 2 + 3 + ( − 5 ) + 7 or ( − 2 ) + ( − 3 ) + 5 + 7 , the prior is the only one with the product of − 2 1 0 .
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Try substituting 1 and -1 for x and get a non-zero result, respectively 4314562560000 and7801587892224. This is a non-problem.
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@A Former Brilliant Member – True, true, it is rather trivial. But should probably be stated in the solution.
See Vieta's Formulae . By factoring the constant term, you get 12 primes, therefore, you now know the absolute values of all the roots. Examination of the coefficient of the x 1 1 term tells you that some of the roots are negative. At this point, you have five attacks on the problem: solving multiple equations in multiple unknowns using Vieta's formulae, guessing which combination of signs allows the sum of the roots to be 19, polynomial division to test each sign possibility, plotting the polynomial to find the approximate roots knowing that the roots are integers and when you get the remaining polynomial to a quartic equation, using closed form solutions.
The roots are {{x->-37},{x->-29},{x->-19},{x->-13},{x->-7},{x->-3},{x->2},{x->5},{x->11},{x->17},{x->23},{x->31}}.
You can see the answer in this plot: