Lösung 3.4:7a
Aus Online Mathematik Brückenkurs 2
(Unterschied zwischen Versionen)
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- | There exists a simple relation between a zero and the polynomial's factorization: | + | There exists a simple relation between a zero and the polynomial's factorization: <math>z=a</math> is a zero if and only if the polynomial contains the factor <math>(z-a)</math>. (This is the meaning of the factor theorem.) |
- | <math>z=a | + | |
- | is a zero if and only if the polynomial contains the factor | + | |
- | <math> | + | |
- | If we are to have a polynomial with zeros at | + | If we are to have a polynomial with zeros at <math>1</math>, <math>2</math> and <math>4</math>, the polynomial must therefore contain the factors <math>(z-1)</math>, <math>(z-2)</math> and <math>(z-4)</math>. For example, |
- | <math>1, | + | |
- | and | + | |
- | <math> | + | |
- | <math> | + | |
- | and | + | |
- | <math> | + | |
+ | {{Displayed math||<math>(z-1)(z-2)(z-4) = z^3-7z^2+14z-8\,\textrm{.}</math>}} | ||
- | <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 14:37, 31. Okt. 2008
There exists a simple relation between a zero and the polynomial's factorization: \displaystyle z=a is a zero if and only if the polynomial contains the factor \displaystyle (z-a). (This is the meaning of the factor theorem.)
If we are to have a polynomial with zeros at \displaystyle 1, \displaystyle 2 and \displaystyle 4, the polynomial must therefore contain the factors \displaystyle (z-1), \displaystyle (z-2) and \displaystyle (z-4). For example,
\displaystyle (z-1)(z-2)(z-4) = z^3-7z^2+14z-8\,\textrm{.} |
Note: It is possible to multiply the polynomial above by a non-zero constant and get another third-degree polynomial with the same roots.