Processing Math: Done
Lösung 3.4:7a
Aus Online Mathematik Brückenkurs 2
(Unterschied zwischen Versionen)
K (Lösning 3.4:7a moved to Solution 3.4:7a: Robot: moved page) |
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- | { | + | There exists a simple relation between a zero and the polynomial's factorization: |
- | < | + | <math>z=a\text{ }</math> |
- | {{ | + | is a zero if and only if the polynomial contains the factor |
+ | <math>\left( z-a \right)</math>. (This is the meaning of the factor theorem.) | ||
+ | |||
+ | If we are to have a polynomial with zeros at | ||
+ | <math>1,\ 2</math> | ||
+ | and | ||
+ | <math>\text{4}</math>, the polynomial must therefore contain the factors | ||
+ | <math>\left( z-1 \right),\ \left( z-2 \right)</math> | ||
+ | and | ||
+ | <math>\left( z-4 \right)</math>. For example, | ||
+ | |||
+ | |||
+ | <math>\left( z-1 \right)\left( z-2 \right)\left( z-4 \right)=z^{3}-7z^{2}+14z-8</math> | ||
+ | |||
+ | |||
+ | NOTE: it is possible to multiply the polynomial above by a non-zero constant and get another third-degree polynomial with the same roots. |
Version vom 15:44, 28. Okt. 2008
There exists a simple relation between a zero and the polynomial's factorization:
z−a
If we are to have a polynomial with zeros at
2
z−1
z−2
z−4
z−1
z−2
z−4
=z3−7z2+14z−8
NOTE: it is possible to multiply the polynomial above by a non-zero constant and get another third-degree polynomial with the same roots.