From Förberedande kurs i matematik 2

Revision as of 15:14, 16 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 3.4:7a moved to Solution 3.4:7a: Robot: moved page) ← Previous diff		Current revision (14:37, 31 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	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.)
-	<center> [[Image:3_4_7a.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	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,
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		+	{{Displayed math||<math>(z-1)(z-2)(z-4) = z^3-7z^2+14z-8\,\textrm{.}</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.

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Current revision

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.