The theorem states: If a polynomial with coefficients a

a

has rational roots p/q, where p and q are integers, then p is a factor of ±a

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.

Consider: x

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.

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.

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.

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

6x

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

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.

q|1 and p|64=1.2

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.

p|12=1.2.2.3

It is always useful to eliminate ±1, and these aren't roots.

Possible p's:

1.1.12 [1]

1.2.6 [2]

1.4.3 [3]

2.2.3 [4]

Possible q's:

1.1.18 [a]

1.3.6 [b]

1.2.9 [c]

2.3.3 [d]

q|18=1.2.3.3

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

The roots are: 1/3, 4/3, 3/2.

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