Lösung 3.4:7a
Aus Online Mathematik Brückenkurs 2
(Unterschied zwischen Versionen)
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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, | 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, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>(z-1)(z-2)(z-4) = z^3-7z^2+14z-8\,\textrm{.}</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. | 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 13:16, 10. Mär. 2009
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.
