Small question about polynomials and factoring them

Alfredo Dawlabany

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Aug 22, 2017
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If we have a polynomial and we want to find a factor of it, I know that we should have a root so that P(x)=(x-a)*g(x) , where "a" is the root and P(x) is the given polynomial and g(x) is an other polynomial. And also I know that one of the divisors of the constant term of a polynomial is a root of it.
The question is that the second method for factoring I mentioned is not always valid. Why ? And when it is valid ?
And also is there a proof for it ?
 

Subhotosh Khan

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If we have a polynomial and we want to find a factor of it, I know that we should have a root so that P(x)=(x-a)*g(x) , where "a" is the root and P(x) is the given polynomial and g(x) is an other polynomial. And also I know that one of the divisors of the constant term of a polynomial is a root of it.
The question is that the second method for factoring I mentioned is not always valid. Why ? And when it is valid ?
And also is there a proof for it ?
Go to:

https://www.google.com/search?q=rational+root+theorem&rlz=1C1GGRV_enUS748US748&oq=rational+root+theorem&aqs=chrome..69i57j0l5.8933j0j8&sourceid=chrome&ie=UTF-8

and read some of the sites mentioned there.
 

ksdhart2

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Mar 25, 2016
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And also I know that one of the divisors of the constant term of a polynomial is a root of it.
The question is that the second method for factoring I mentioned is not always valid. Why ? And when it is valid ?
To me, this sounds like you're talking about the Rational Root Theoremhttp://www.purplemath.com/modules/rtnlroot.htm, although you appear to have some misconceptions about how it works. Obviously, this theorem can only apply if the constant term has integer divisors, so right there you're restricted to a very small subset of polynomials (although here "very small" may seem like a bit of a misnomer, as there's still infinitely many such polynomials). The rational root theorem also depends not just on the divisors of the constant term, but also of the coefficient of the highest power. Further, the rational root theorem isn't guaranteed to generate a root, only a list of numbers that might be roots., and even if it does generate one or more roots, it probably won't generate all the roots of an arbitrary polynomial. As its name indicates, it can only ever find rational roots. To find any irrational or complex roots, you'll need other methods.

As an example, take the polynomial \(\displaystyle 2x^3+3x^2+4x+16\). If we apply the rational root theorem, we see that the possible rational roots are \(\displaystyle \pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \text{ and } \pm \dfrac{1}{2}\). However, checking these roots reveals that none of them are actually roots. In fact, the one real root is at \(\displaystyle x \approx 2.2211\) and the other two roots are complex.
 

JeffM

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Sep 14, 2012
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I want to supplement the previous answer.

The rational root theorem gives you the list of every RATIONAL number that MAY BE a zero of a polynomial in one unknown with INTEGER coefficients. It does not say that there is such a zero.

A special version applies if the coefficient of the highest order term is 1. Then you can apply the integer root theorem, which gives you the list of every INTEGER that MAY BE a zero of a polynomial in one unknown with INTEGER coefficients. Again, it does not say that there is such a zero.

For an in depth discussion, start with https://en.m.wikipedia.org/wiki/Rational_root_theorem
 
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