SOLUTION: Solve x^3 - x^2 + 5x - 7 = 0 all possible roots 1, -1, 7, -7 Using synthetic division none of them work tried it several times! 1 1 -1 5 -7

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Question 1149598: Solve x^3 - x^2 + 5x - 7 = 0 all possible roots 1, -1, 7, -7 Using synthetic division none of them work tried it several times!
1 1 -1 5 -7
Found 3 solutions by Alan3354, Edwin McCravy, ikleyn:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Solve x^3 - x^2 + 5x - 7 = 0 all possible roots 1, -1, 7, -7 Using synthetic division none of them work tried it several times!
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None of those are roots.
There's 1 real root, not rational. ~ 1.29906257
Answer by Edwin McCravy(20064)   (Show Source): You can put this solution on YOUR website!
x^3 - x^2 + 5x - 7 = 0 
It has no rational roots.  But you might be interested to know that it comes close
to having root 1.3. It even works if you round everything off to 1 decimal place.

1.3 | 1    -1    5     -7
    |     1.3    0.4    7.0 
      1   0.3    5.4    0

But that's not really solving it.  It's just for fun.  Don't expect all polynomials to have rational roots.

Edwin
Answer by ikleyn(52879)   (Show Source): You can put this solution on YOUR website!
.

If you try to apply the Rational root theorem, then you should know EXACTLY and PRECISELY what it states.



    It states that IF the polynomial has rational roots, THEN these roots are among the certain set of numbers  { +/-  }.



But this theorem DOES NOT state that the polynomial MUST have the roots among this set.


The polynomial may HAVE NO rational roots, at all.


It is exactly the case with the given polynomial.



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Comment from student: Thank you for clarifying it has no rational roots but it has an irrational root about 1.3
how do you get that algebraically...without using a graphing calculator?
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My response : You may use the Cardano's formula for solution of the polynomial equation of the degree 3.

https://en.wikipedia.org/wiki/Cubic_equation



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