# polynomial powers and root quatity

• Aug 14th 2012, 09:40 AM
anthonye
polynomial powers and root quatity
Hi;
I thought that if I had a 4degree polynomial it has 4 roots,
a 5degree polynomial has 5 roots ect...

But I'm just starting to read an article on Descartes' rule of signs and it says
you can dertermine the amount of roots a polynomial can have using this rule,

why have the rule if we can tell from the power?
• Aug 14th 2012, 09:58 AM
emakarov
Re: polynomial powers and root quatity
Quote:

Originally Posted by anthonye
I thought that if I had a 4degree polynomial it has 4 roots,
a 5degree polynomial has 5 roots ect...

This is true (by the fundamental theorem of algebra) only if you allow complex roots. The polynomial x⁴ + 1 has no real roots because x⁴ ≥ 0 for all x.

Besides, Descartes' rule of signs allows estimating the number of positive roots. For example, (x + 1)(x - 1) has two roots but only one positive root.
• Aug 14th 2012, 10:01 AM
anthonye
Re: polynomial powers and root quatity
So if I account for complex roots the higest power is the number of roots?

Descartes' only allows for positive roots ?
• Aug 14th 2012, 10:03 AM
skeeter
Re: polynomial powers and root quatity
Quote:

Originally Posted by anthonye
So if I account for complex roots the higest power is the number of roots?

Descartes' only allows for positive roots ?

Descartes' Rule of Signs
• Aug 14th 2012, 01:53 PM
HallsofIvy
Re: polynomial powers and root quatity
DesCarte's rule of signs says that the number of positive real roots of a polynomial is either equal to the number of changes of sign or less than that by a multiple of two. To deal with negative roots, replace "x" with "-x".

For example, \$\displaystyle x^5+ 4x^4- x^3+ x^2+ x- 5\$ has 3 changes of sign: from positive to negative, at \$\displaystyle -x^3\$, back to positive at \$\displaystyle + x^2\$ and negative again at -5. That tells us that there are either 3 or 1 postive roots. If we replace x with -x, the odd powers of x change sign giving \$\displaystyle -x^5+ 4x^4+ x^3+ x^2- x- 5\$ which has 2 changes in sign and so either 2 or 0 negative roots. Of course, that does NOT say anything about complex roots. DesCarte's rule of signs tells us there may be
1) 3 positive, 2 negative, and 0 complex roots
2) 1 positive, 2 negative, and 2 complex roots
3) 3 positive, 0 negative, and 0 complex roots
4) 1 positive, 0 negative, and 4 complex roots
There have to be an even number of complex roots, of course, because, since the coefficients are real numbers, the complex roots come in complex conjugate pairs.