Lösung 3.4:7a

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There exists a simple relation between a zero and the polynomial's factorization:
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<center> [[Image:3_4_7a.gif]] </center>
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<math>z=a\text{ }</math>
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is a zero if and only if the polynomial contains the factor
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<math>\left( z-a \right)</math>. (This is the meaning of the factor theorem.)
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If we are to have a polynomial with zeros at
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<math>1,\ 2</math>
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and
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<math>\text{4}</math>, the polynomial must therefore contain the factors
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<math>\left( z-1 \right),\ \left( z-2 \right)</math>
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and
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<math>\left( z-4 \right)</math>. For example,
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<math>\left( z-1 \right)\left( z-2 \right)\left( z-4 \right)=z^{3}-7z^{2}+14z-8</math>
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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: \displaystyle z=a\text{ } is a zero if and only if the polynomial contains the factor \displaystyle \left( z-a \right). (This is the meaning of the factor theorem.)

If we are to have a polynomial with zeros at \displaystyle 1,\ 2 and \displaystyle \text{4}, the polynomial must therefore contain the factors \displaystyle \left( z-1 \right),\ \left( z-2 \right) and \displaystyle \left( z-4 \right). For example,


\displaystyle \left( z-1 \right)\left( z-2 \right)\left( z-4 \right)=z^{3}-7z^{2}+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.