The equation calls for you to factor using the P over Q method, and I tried every single "possible answer" with the +/- 1, etc.... method and NONE of the possible answers were correct, because I did not get a 0 at the end of the synthetic devisions..... so would the answer be that it is unfactorable ?
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Your strategy would indicate that there are no rational roots
[± p/q from the rational root theorem]
If your calculations are correct,
then the polynomial can not be factored with real, rational coefficients.
@ß
[± p/q from the rational root theorem]
If your calculations are correct,
then the polynomial can not be factored with real, rational coefficients.
@ß
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The Rational Root Theorem does not guarantee that there is a rational solution. So, there are times when none of the possible solutions will work. The equation will have a solution, it just won’t be rational.
Edit: But you need to check the possible zeros by plugging them into your function to see if they make it zero. i.e.: f(a) = 0. Then you use synthetic division to find the other factors.
Edit: But you need to check the possible zeros by plugging them into your function to see if they make it zero. i.e.: f(a) = 0. Then you use synthetic division to find the other factors.
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How are we supposed to tell if an equation is non-factorisable if we can't see the equation ? Telepathy ?