- 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

- p is even, and q is odd
- p is odd and q is even
- p is odd and q is odd

- 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.

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)

Multiplying throughout by q gives:

p

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

Back to the equation.

In case 1, p is even, and q is odd, we have:

p

So p may be even and q odd.

In case 2, p is odd and q is even:

p

So, case 2 isn't possible for this equation.

In case 3, p is odd and q is odd:

p

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

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.

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

p

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

p

This is not possible, because an odd number would equal an even number.

In case 2:

p

This is not possible.

In case 3:

p

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.

We substitute x=p/q and multiply throughout by q

p

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

p

This is contraditory, so case 1 is not possible.

In case 2

p

Again, not possible.

In case 3:

p

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)

According to the Rational Root Theorem, p|2=1.2 and q|24=1.2

We substitute x=p/q and multiply throughout by q

24p

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

24p

e+e+e+e=e, 0 is even, so this is possible.

p|2=±1.2 and q|24=±1.2

So the possible roots are {±2/1,±2/3}

case 2:

24p

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

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.

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

According to the Rational Root Theorem, p|191=1.191 and q|1000=2

We substitute x=p/q and multiply throughout by q

1000p

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

1000p

e+e+e+o=o, 0 is even, so this is impossible.

case 2:

1000p

e+e+e+e=e, 0 is even, so this is possible.

So p|±1.191, and q|2

case 3:

1000p

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

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