I have been told that to find factors of a polynomial (nth degree) we have to find the factors of constant term and that of coefficient of leading term of the polynomial in concern.

The possible integral zeros of the polynomial will be from the factor set of the constant term while rational zeros would be from set of each factor of constant / each factor of coefficient of the leading term.

Now I have to replace one by one value of $X$ for each integral and rational factors founded above and check if the polynomial results to $0$ (zero). The issue is that I go into deep / lengthy calculation if suppose constant term is $140$ and coefficient of leading term is $6$ per say.

Factors of $140 = -1,+1,-2,+2,-4,+4,-5,+5,-7,+7,.....$

Factors of $6 = -1,+1,-2,+2,-3,+3,-6,+6 $

Integral roots of the polynomial (set range) = $-1,+1,-2,+2,-4,+4,-5,+5,-7,+7,.....$

Rational roots of the polynomial (set range) = $(-1,+1,-2,+2,-4,+4,-5,+5,-7,+7,.....)/ (-1,+1,-2,+2,-3,+3,-6,+6 )$

Taking one by one value and testing for zero is a very lengthy time consuming method - is there a quick easy way to find zeros of the pronominal?

  • $\begingroup$ In general, finding the rational roots of a polynomial requires plugging in all of these values. Some people automate the process using synthetic division. There are a few tricks in synthetic division that allow you to exclude possible values quickly, such as if the values on the bottom row are always positive. $\endgroup$ – Michael Burr Jul 14 '15 at 9:07
  • $\begingroup$ The quick easy way to find zeros of a polynomial is to use the appropriate command in Maple or Mathematica or some other computer algebra package. Doing it by hand, I'm afraid that for every one of those rational numbers you refer to, there's a polynomial with constant term 140 and leading coefficient 6 having that rational solution and no other rtational solution. $\endgroup$ – Gerry Myerson Jul 14 '15 at 10:39

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