The rational root theorem, or Rational Root Test can be used to do a
quick check whether a
polynomial has rational roots. This page is about using the theorem and
not about the theory. The theorem states: If a polynomial with coefficients an,
an-1...a0 : anxn+a1xn-1...a0=0 has rational roots p/q, where p and q are integers, then p is a factor
of ±a0 and q is a factor of
or using "|" to mean is a factor of, p|a0 and q|an.
According to the Rational Root Theorem, the roots can be plus or minus,
although p|a0 implies
that p can be plus or minus (because, for instance, both 2 and -2 are
factors of 4, so writing ± serves simply as a reminder of
this.). The numbers p and q have no common factors, they are mutually prime. If
this were not so, the original equation would not be in normal form,
and therefore simplified. Also Descartes Rule of
Signs might tell us that all the roots are positive, or
negative, when the options from the Rational Root Theorem are reduced.
We should not ashamed to give trivial examples! Consider: x2-3x+2=0 If this has rational roots p/q (which we know it has), then p is a
factor of 2 and q is a factor of 1. So the roots are some combination of the factors in the fraction below:
We must use all the numbers, that is we need to use both 1 and 2 to
make possible roots. Where all the roots are rational, we cannot use
just the "1" as a numerator. We need to use the 2 also. So 1 and 1
won't do as numerators, although 2 and 2 are possible (because the 1 is
always automatically included in any integer.) We test each of the roots in the original equation and, in this case,
we find x=1/1 or 2/1, which are 1 and 2, of course.
x2+2x-15 If this has rational roots, then they are factors of:
We must use all the numbers, so we cannot take 1 and 3 as possible
numerators, because we must include 5 somewhere. So if 1 is one
numerator, then 15 must be the other (because there are 2 roots and we
need to use up all the numbers.) By knowing or trying, we find -5 and 3 are the required roots.
2x2-3x+5 If this has rational roots, then they are factors of:
We can try the combinations 1/2, 1/1, 5/1, 5/2 both plus and minus (but
Descartes Rule of Signs
tells us all the roots are positive if they are real, so we need to
check positive roots only.), but
we fail to find any roots. The roots of this are approximately: x0=0.75+1.3919i x1=0.75 – 1.3919i And these aren't rational numbers. So, as expected, when the roots
aren't rational, they do not conform with the rational root theorem.
21x2+11x-2 If this has rational roots, then they are factors of:
So, if there are rational roots, they are among: 1/1, 1/3, 1/7, 2/1,
2/3, 2/7 Because we need to use up all the factors, so for example, if one root
is |1/3|, then the other is |2/7|, etc..
In fact they are 1/7 and -2/3
6x3-73x2-86x+273 If this has rational roots, then they are factors of:
Because the roots p/q are such that p and q have no common factors,
then ±3/3 isn't an option for a root. Also we need to use up
all the numbers, so if 3 is a numerator, the other two numerators must
be 7 and 13 or 1 and 91. Either way the numbers are used up. In fact, 3/2, -7/3 and 13 satisfy the equation
x5+3x4+3x3+3x2+3x+2 If this has rational roots, they are factors of ±(1.2)/(1).
Of the 4 combinations, only -2 is a root.
In fact, the roots are: Root0= – 2 Root1=0.3090169944+0.9510565163i Root2= – 0.8090169944+0.5877852523i Root3= – 0.80901699445 – 0.5877852523i Root4=0.3090169944 – 0.9510565163i So, it has only one rational root, which we found.
x3+14x2+56x+64 q|1 and p|64=1.26 p/q divides (1.26)/1 If there are rational roots, then there denominators are all 1 (because
that is the only factor we have for the numerators). The numbers are all whole numbers. We can quickly eliminate ±1, by substituting in the equation. The possible numerators include: 2.2.16, 2.4.8, 4.4.4 2 occurs in 2 of the options, as does 4. So ±2 and
±4 are worth checking. In fact, -2, -4 and -8 are the roots of the equation.
It is always useful to eliminate ±1, and these aren't roots.
In the p's, 3 can be the numerator of 1 or 2 only, so possible roots are: ±3,±3/2.
Also in the p's 2 can be the numerator of 1 or 3 only