The following rules can sometimes be helpful. Say o is an odd number
and e is an even number. We can sometimes eliminate some possibilities
given by the Rational
Root Theorem, and save on unnecessary applications of Horner's Scheme.
e+e=e (That is, adding two even numbers gives an even
number)
e+o=o (Adding an even and an odd number gives an odd number)
o+o=e
o*e=e;
o*o=o
o^{n}=o (n is a natural number)
e^{n}=e
For a polynomial, if it has rational roots, p/q, (p and q are integers)
then there are three possibilities in terms of odd and even:
p is even, and q is odd
p is odd and q is even
p is odd and q is odd
The
option p is even and q is even is not present when the equation is
normalised, because cases where 2p/2q might exist become p/q, because
the common factor cancels. That is, p and q are relatively prime, they
share no common factors.
If testing an equation with these rules produces a
contracictory
result, then that combination of odd and even is impossible. We are
sure such roots do not exist.
If the results are logical, then it means that combination
of odd and even is not illogical, but it does notprove that such roots exist. One cannot
use logic to prove existence.
Example 1
Once again, do not be contemptuous of trivial examples! x^{2}+6x+8=0
By the Rational Root
Theorem, if p/q are roots of the equation then p|8 and q|1.
Substituting x=p/q in the equation gives:
(p/q)^{2}+6p/q+8=0
Multiplying throughout by q gives:
p^{2}+6pq+8q^{2}=0 (This can
just be written down from the original equation, after practice)
For the odd and even nature of roots, (repeating the above) there are
three possibilites:
p is even, and q is odd
p is odd and q is even
p is odd and q is odd
The option p is even and q is even is not present when the equation is
normalised, because cases where 2p/2q might exist become p/q, because
the common factor cancels. That is, p and q are relatively prime, that
is, they share no common factors.
Back to the equation.
In case 1, p is even, and
q is odd, we have:
p^{2 }(even)+6pq(even)+8q^{2}
(even)=0 (even)
So p may be even and q odd.
In case 2, p is odd and q
is even:
p^{2 }(odd)+6pq(even)+8q^{2}
(even)=o+e+e=o=0 (even) [contradictory]
So, case 2 isn't possible for this equation.
In case 3, p is odd and
q is odd:
p^{2 }(odd)+6pq(even)+8q^{2}
(even)=o+e+e=o=0 (even) [contradictory]
So p and q cannot both be odd.
Therefore p is even and q is odd.
From above, the rational root theorem tells us that:
p|8 and q|1
Naturally, there is only one option for q, which is 1, an odd number
p|8=1.2^{3}, and we can only eliminate 1.
Descartes Rule of Signs tells us all the roots are negative.
The possible roots are x={-2, -4, -8}, if there are rational roots.
Actually, the roots are -2, -4.
Example 2
x^{2}+x+1=0
If there are rational roots, then the roots can be written as p/q,
where, as usual, p and q are integers.
Substituting x=p/q in the equation and multiplying throughout by q^{2}
p^{2}+pq+q^{2}=0
For the odd and even nature of roots, (repeating the above) there are
three cases:
p is even, and q is odd
p is odd and q is even
p is odd and q is odd
In case 1:
p^{2} (even)+pq (even)+q^{2}(odd)=e+e+o=0
(e)
This is not possible, because an odd number would equal an even number.
In case 2:
p^{2} (odd)+pq (even)+q^{2}(even)=o+e+e=0
(e)
This is not possible.
In case 3:
p^{2} (odd)+pq (odd)+q^{2}(odd)=o+o+o=0
(e)
This is contradictory and therefore not possible.
The rational roots of the equation are no combination of odd and even,
and therefore such roots do not exist.
Using the quadratic formula,
we can determine the roots to be:
-1/2 ±√(3)/2i
Which confirms that there are no rational roots for this equation.
Example 3
x^{3}+6x^{2}+2x+1
We substitute x=p/q and multiply throughout by q^{3},
writing the result down:
p^{3}+6p^{2}q+2pq^{2}+q^{3}=0
or the odd and even nature of roots, (repeating the above) there are
three cases:
p is even, and q is odd
p is odd and q is even
p is odd and q is odd
In case 1:
p^{3} (even)+6p^{2}q (even)+2pq^{2}
(even)+q^{3} (odd)=0 (even)
This is contraditory, so case 1 is not possible.
In case 2
p^{3} (odd)+6p^{2}q (even)+2pq^{2}
(even)+q^{3} (even)=0 (even)
Again, not possible.
In case 3:
p^{3} (odd)+6p^{2}q (even)+2pq^{2}
(even)+q^{3} (odd)=0 (even)
This is logical, so the equation could have a root where p and q are
both odd.
By Descartes Rule of
Signs, all the real roots are negative.
So the possible root, by the Rational Root Theorem is -1.
This isn't a root of the equation. So while such a root is not impossible, it
doesnot exist.
The roots of the equation are:
Cubic Formula Solution (errors include rounding errors):
x0= – 5.678823030129 (Error=1.7e – 11)
x1= – 0.160588484935+0.387690811418i (Error=0)
x2= – 0.160588484935 – 0.387690811418i (Error=0)
Example 4
24x^{3}+2x^{2}+-9x-2
According to the Rational Root Theorem, p|2=1.2 and q|24=1.2^{3}.3
We substitute x=p/q and multiply throughout by q^{3},
writing the result down:
24p^{3}+2p^{2}q-9pq^{2}-2q^{3}=0
or the odd and even nature of roots, (repeating the above) there are
three cases:
p is even, and q is odd
p is odd and q is even
p is odd and q is odd
case 1:
24p^{3} (even)+2p^{2}q
(even)-9pq^{2} (even)-2q^{3}
(even)=0
e+e+e+e=e, 0 is even, so this is possible.
p|2=±1.2 and q|24=±1.2^{3}.3, from above. So p=±2
and q=±1, or ±3
So the possible roots are {±2/1,±2/3}
case 2:
24p^{3} (even)+2p^{2}q
(even)-9pq^{2} (even)-2q^{3}
(even)=0
e+e+e+e=e, 0 is even, so this is possible.
So p=±1, and q={±2,±4±,6,±8}, giving ±(1/2, 1/4, 1/6, 1/8) case 3:
24p^{3} (even)+2p^{2}q
(even)-9pq^{2} (odd)-2q^{3}
(even)=0
e+e+o+e=o, 0 is even, so this case isn't possible.
Looking at the graph, the roots appear to be 2/3, -1/4, -1/2
And this being successful, checking -1/4. The last line is the equation
24x+12=0, which verifies the last root as -1/2.
Example 5
1000x^{3}-1254x^{2}-496x+191
According to Descartes,
this
equation has 2 positive real roots (or none), and one negative real
root. Actually, the graph below shows this to be true. Vietas Formulas
tell us:
The sum of the roots=1254/1000, and
The product is -191/1000
1000x^{3}-1254x^{2}-496x+191
According to the Rational
Root Theorem, p|191=1.191 and q|1000=2^{3}.5^{3}
We substitute x=p/q and multiply throughout by q^{3},
writing the result down:
1000p^{3}-1254p^{2}q-496pq^{2}+191q^{3}
or the odd and even nature of roots, (repeating the above) there are
three cases:
p is even, and q is odd
p is odd and q is even
p is odd and q is odd
case 1:
1000p^{3}(even)-1254p^{2}q(even)-496pq^{2}(even)+191q^{3}(odd)
=0(even)
e+e+e+o=o, 0 is even, so this is impossible.
case 2:
1000p^{3}(even)-1254p^{2}q(even)-496pq^{2}(even)+191q^{3}(even)
=0(even)
e+e+e+e=e, 0 is even, so this is possible.
So p|±1.191, and q|2^{3}.5^{3},
giving ±(1/2, 1/4, 1/8, 191/2, 191/4, etc) case 3:
1000p^{3}(odd)-1254p^{2}q(even)-496pq^{2}(even)+191q^{3}(odd)
=0(even)
o+e+e+o=e, 0 is even, so this case is also possible.
p/q could be ± (1/5, 1/25, 191/5, etc)
Looking at the graph, the roots appear to be -1/2, 1/4, 1.5.
However, in our analysis above, 1.5 does not occur as a
posible rational root.
The sum of the roots=1254/1000, and
The product is -191/1000
Hower, (-1/2)(1/4)(1.5)=-3/16=0.185, and
-1/2+1/4+3/2=1 1/4=1.25
The roots for
Cubic: 1000x³–1254x²–496x+191 are:
x1=1.4997993055 (Remainder=0)
x2=–0.5003313644 (Remainder=0)
x3=0.254532059 (Remainder=0)
The remainder is the result of substituting the value in the equation,
rounded to 10 decimal places