Ken Ward's Mathematics Pages

Even and Odd Roots

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.

1. e+e=e (That is, adding two even numbers gives an even number)
2. e+o=o (Adding an even and an odd number gives an odd number)
3. o+o=e
4. o*e=e;
5. o*o=o
6. oⁿ=o (n is a natural number)
7. eⁿ=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:

1. p is even, and q is odd
2. p is odd and q is even
3. 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 not prove that such roots exist. One cannot use logic to prove existence. 

Example 1

Once again, do not be contemptuous of trivial examples! x²+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:
p²+6pq+8q²=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:

1. p is even, and q is odd
2. p is odd and q is even
3. 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:
p² (even)+6pq(even)+8q² (even)=0 (even)
So p may be even and q odd.
In case 2, p is odd and q is even:
p² (odd)+6pq(even)+8q² (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:
p² (odd)+6pq(even)+8q² (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=1.2³, 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

x²+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 q²
p²+pq+q²=0
For the odd and even nature of roots, (repeating the above) there are three cases:

1. p is even, and q is odd
2. p is odd and q is even
3. p is odd and q is odd

In case 1:
p² (even)+pq (even)+q²(odd)=e+e+o=0 (e)
This is not possible, because an odd number would equal an even number.

In case 2:
p² (odd)+pq (even)+q²(even)=o+e+e=0 (e)
This is not possible.

In case 3:
p² (odd)+pq (odd)+q²(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

x³+6x²+2x+1
We substitute x=p/q and multiply throughout by q³, writing the result down:
p³+6p²q+2pq²+q³=0
or the odd and even nature of roots, (repeating the above) there are three cases:

1. p is even, and q is odd
2. p is odd and q is even
3. p is odd and q is odd

In case 1:
p³ (even)+6p²q (even)+2pq² (even)+q³ (odd)=0 (even)
This is contraditory, so case 1 is not possible.

In case 2
p³ (odd)+6p²q (even)+2pq² (even)+q³ (even)=0 (even)
Again, not possible.

In case 3:
p³ (odd)+6p²q (even)+2pq² (even)+q³ (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 does not 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

24x³+2x²+-9x-2
According to the Rational Root Theorem, p|2=1.2 and q|24=1.2³.3
We substitute x=p/q and multiply throughout by q³, writing the result down:
24p³+2p²q-9pq²-2q³=0
or the odd and even nature of roots, (repeating the above) there are three cases:

1. p is even, and q is odd
2. p is odd and q is even
3. p is odd and q is odd

case 1:
24p³ (even)+2p²q (even)-9pq² (even)-2q³ (even)=0
e+e+e+e=e, 0 is even, so this is possible.
p|2=±1.2 and q|24=±1.2³.3, from above. So p=±2 and q=±1, or ±3
So the possible roots are {±2/1,±2/3}

case 2:
24p³ (even)+2p²q (even)-9pq² (even)-2q³ (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:
24p³ (even)+2p²q (even)-9pq² (odd)-2q³ (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



Checking 2/3 with Horner's Scheme:

	24	2	-9	-2
2/3	0	16	12	2
	24	18	3	0
-1/4	0	-6	-3
	24	12	0

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

1000x³-1254x²-496x+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


1000x³-1254x²-496x+191
According to the Rational Root Theorem, p|191=1.191 and q|1000=2³.5³
We substitute x=p/q and multiply throughout by q³, writing the result down:
1000p³-1254p²q-496pq²+191q³
or the odd and even nature of roots, (repeating the above) there are three cases:

1. p is even, and q is odd
2. p is odd and q is even
3. p is odd and q is odd

case 1:
1000p³(even)-1254p²q(even)-496pq²(even)+191q³(odd) =0(even)
e+e+e+o=o, 0 is even, so this is impossible.

case 2:
1000p³(even)-1254p²q(even)-496pq²(even)+191q³(even) =0(even)
e+e+e+e=e, 0 is even, so this is possible.
So p|±1.191, and q|2³.5³, giving ±(1/2, 1/4, 1/8, 191/2, 191/4, etc)
case 3:
1000p³(odd)-1254p²q(even)-496pq²(even)+191q³(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







Ken Ward's Mathematics Pages

---

---

Faster Arithmetic - by Ken Ward