Solving polynomial equations, descartes theorem

A large part of the algebra courses at upper secondary-school level are devoted to solving equations or factorization of polynomials. This is often the same thing.

Some terminology. All of these are the same:
Solving a polynomial equation.
Finding roots of a polynomial equation p(x)=0.
Finding zeros of a polynomial equation p(x)
Factorizing a polynomial function p(x).

There is a factor for every root and vice versa.
(x-r) is a factor if and only if r is a root according to the Factor theorem.

How to solve equations step-by-step:
1. If solving an equation, put it in standard form with 0 on one side and simplif.
2. Know hoewmany roots to expect.
3. Find one factor or root. (Several techniques available)
4. Divide by your factor. This leaves you with a new reduced polynomial. whose degree is 1 less. For the rest of the problem you’ll work with the reduced polynomial and not the original.
5. Now you have a quadratic or linear equation which you already know how to solve.
Write down the solution.

There is no general solution for solving equations of degree 5 and higher.
Try to factorize the polynomial as much as possible.
Equations of degree four or less can be solved by stanard methods.

A plynomial of degree n will have n roots some of which may be multiple roots. This is according to the Fundamental theorem of Algebra.

Descartes Rule of Sign:
Tells you the how many positiv or negative real zeroes the polynomial has.
1. The number of positive roots of p(x)=0 is either equal to the number of variations in sign of p*(x) or less than that by an even numer.
2. The number of negative roots of p(x) = 0 is either equal to the number of variations in sign of p(-x)=0 or less than that by an even number.

Complex Roots

If a polynomial has real coefficients, then either all roots are real or there are an even number of non-real complex roots, in conjugate pairs.

For example, if 5+2i is a zero of a polynomial with real coefficients, then 5−2i must also be a zero of that polynomial. It is equally true that if (x−5−2i) is a factor then (x−5+2i) is also a factor.

This is true because when you have a factor with an imaginary part and multiply it by its complex conjugate you get a real result:

(x−5−2i)(x−5+2i) = x²−10x+25−4i² = x²−10x+29

If (x−5−2i) was a factor but (x−5+2i) was not, then the polynomial would end up with imaginaries in its coefficients, no matter what the other factors might be. If the polynomial has only real coefficients, then any complex roots must occur in conjugate pairs.


Om mattelararen

Licentiate of Philosophy in atomic Physics Master of Science in Physics
Det här inlägget postades i Algebra, Gymnasiematematik(high school math), matematik 2c och har märkts med etiketterna , , , , . Bokmärk permalänken.


Fyll i dina uppgifter nedan eller klicka på en ikon för att logga in: Logo

Du kommenterar med ditt Logga ut / Ändra )


Du kommenterar med ditt Twitter-konto. Logga ut / Ändra )


Du kommenterar med ditt Facebook-konto. Logga ut / Ändra )

Google+ photo

Du kommenterar med ditt Google+-konto. Logga ut / Ändra )

Ansluter till %s