3.4 Komplexe Polynome

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{{Vald flik|[[3.4 Komplexa polynom|Teori]]}}
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{{Vald flik|[[3.4 Komplexa polynom|Theory]]}}
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{{Ej vald flik|[[3.4 Övningar|Övningar]]}}
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{{Ej vald flik|[[3.4 Övningar|Exercises]]}}
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'''Innehåll:'''
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'''Content:'''
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* Faktorsatsen
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* Factor theorem.
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* Polynomdivision
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* Polynomial long division
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* Algebrans fundamentalsats
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* Fundamental theorem of algebra
}}
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{{Info|
{{Info|
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'''Lärandemål:'''
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'''Learning outcomes:'''
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Efter detta avsnitt ska du ha lärt dig att:
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After this section, you will have learned to:
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* Utföra polynomdivision.
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* Perform polynomial division.
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* Förstå sambandet mellan faktorer och nollställen till polynom.
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* Understand the relationship between factors and zeros of polynomials.
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* Veta att en polynomekvation av grad ''n'' har ''n'' rötter (räknade med multiplicitet).
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* Know that a polynomial equation of degree n has n roots (including multiplicity).
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* Veta att reella polynomekvationer har komplexkonjugerade rötter.
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* Know that real polynomial equations have complex conjugate roots.
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}}
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== Polynom och ekvationer ==
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== Polynom and equations ==
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Ett uttryck på formen
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An expression of the form
{{Fristående formel||<math>a_nx^n+a_{n-1}x^{n-1} + \ldots + a_2x^2 + a_1x+a_0</math>}}
{{Fristående formel||<math>a_nx^n+a_{n-1}x^{n-1} + \ldots + a_2x^2 + a_1x+a_0</math>}}
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där <math>n</math> är ett naturligt tal, kallas ett ''polynom'' av grad <math>n</math> i en obestämd variabel <math>x</math>. Talet <math>a_1</math> kallas koefficienten för <math>x</math>, <math>a_2</math> koefficienten för <math>x^2</math>, etc. Konstanten <math>a_0</math> kallas ''konstanttermen''.
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where <math>n</math> is an integer, is called a ''polynomial'' of degree <math>n</math> in an unknown variable <math>x</math>. The number <math>a_1</math> is called the coefficient of <math>x</math>, <math>a_2</math> the coefficient of <math>x^2</math>, etc. The constant <math>a_0</math> is called the ''constant term''.
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Polynom är grundläggande för en stor del av matematiken och visar bl.a. upp stora likheter med våra heltal, vilket gör att vi kan räkna med polynom på liknande sätt som med heltalen.
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Polynomials are essential for a large part of mathematics and have many properties which display great similarities with our integers, which means that we can do calculations with polynomials in a similar way as with integers.
<div class="exempel">
<div class="exempel">
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'''Exempel 1'''
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''' Example 1'''
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Jämför följande heltal skrivet i basen 10,
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Compare the following integer written using a base 10,
{{Fristående formel||<math>1353= 1\cdot 10^3 + 3\cdot 10^2 + 5\cdot 10 + 3</math>}}
{{Fristående formel||<math>1353= 1\cdot 10^3 + 3\cdot 10^2 + 5\cdot 10 + 3</math>}}
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med ett polynom i <math>x</math>
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with the polynomial in <math>x</math>
{{Fristående formel||<math>x^3 + 3x^2 + 5x + 3 = 1\cdot x^3 + 3\cdot x^2 + 5\cdot x + 3</math>}}
{{Fristående formel||<math>x^3 + 3x^2 + 5x + 3 = 1\cdot x^3 + 3\cdot x^2 + 5\cdot x + 3</math>}}
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och sedan följande divisioner,
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and then the following divisions,
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*<math>\quad\frac{1353}{11} = 123 \qquad</math> eftersom <math>\ 1353= 123\cdot 11\,</math>,
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*<math>\quad\frac{1353}{11} = 123 \qquad</math> as <math>\ 1353= 123\cdot 11\,</math>,
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*<math>\quad\frac{x^3 + 3x^2 + 5x + 3}{x+1} = x^2+2x+3\qquad</math> eftersom <math>\ x^3 + 3x^2 + 5x + 3= (x^2+2x+3)(x+1)\,</math>.
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*<math>\quad\frac{x^3 + 3x^2 + 5x + 3}{x+1} = x^2+2x+3\qquad</math> since <math>\ x^3 + 3x^2 + 5x + 3= (x^2+2x+3)(x+1)\,</math>.
</div>
</div>
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Om <math>p(x)</math> är ett polynom av grad <math>n</math> så kallas <math>p(x)=0</math> en ''polynomekvation'' av grad <math>n</math>. Om <math>x=a</math> är ett tal sådant att <math>p(a)=0</math> så kallas <math>x=a</math> en ''rot'', eller lösning till ekvationen. Man säger också att <math>x=a</math> är ett ''nollställe'' till <math>p(x)</math>.
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If <math>p(x)</math> is a polynomial of degree <math>n</math> then <math>p(x)=0</math> is called a ''polynomial equation'' of degree <math>n</math>. If <math>x=a</math> is a number such that <math>p(a)=0</math> then <math>x=a</math> is called a ''root'', or solution to the equation. One also says that <math>x=a</math> is a ''zero'' of <math>p(x)</math>.
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Som exemplet ovan visade kan polynom divideras precis som heltal. En sådan division går, precis som för heltal, i allmänhet inte jämnt upp. Om t.ex. <math>37</math> divideras med <math>5</math>, får man
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As the above example showed polynomial, can be divided just like integers. Polynomial division, just like integer division usually is not exact. If, for example, <math>37</math> is divided by <math>5</math>, one gets
{{Fristående formel||<math>\frac{37}{5} = \frac{35+2}{5}=7+\frac{2}{5}\,\mbox{.}</math>}}
{{Fristående formel||<math>\frac{37}{5} = \frac{35+2}{5}=7+\frac{2}{5}\,\mbox{.}</math>}}
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Uträkningen kan även skrivas <math>\ 37= 7\cdot 5+2\,</math>. Talet 7 kallas ''kvot'' och talet 2 ''rest''. Man säger att division av 37 med 5 ger kvoten 7 och resten 2.
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The calculation can also be written as <math>\ 37= 7\cdot 5+2\,</math>. The number 7 is called the ''quotient'' and the number 2 the ''remainder''. One says that dividing 37 by 5 gives the quotient 7 and the remainder 2.
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Om <math>p(x)</math> och <math>q(x)</math> är polynom så kan man på liknande sätt dividera <math>p(x)</math> med <math>q(x)</math> och entydigt bestämma polynom <math>k(x)</math> och <math>r(x)</math> så att
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If <math>p(x)</math> and <math>q(x)</math> are polynomials one similarly can divide <math>p(x)</math> with <math>q(x)</math> and unambiguously determine polynomials <math>k(x)</math> and <math>r(x)</math> such that
{{Fristående formel||<math>\frac{p(x)}{q(x)} = k(x)+ \frac{r(x)}{q(x)}\,\mbox{,}</math>}}
{{Fristående formel||<math>\frac{p(x)}{q(x)} = k(x)+ \frac{r(x)}{q(x)}\,\mbox{,}</math>}}
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eller <math>\ p(x)= k(x)\cdot q(x)+r(x)\,</math>. Man säger här att polynomdivisionen ger kvoten <math>k(x)</math> och resten <math>r(x)</math>.
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or <math>\ p(x)= k(x)\cdot q(x)+r(x)\,</math>. One here says that polynomial division has resulted in a quotient <math>k(x)</math> and remainder <math>r(x)</math>.
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Det är uppenbart att en division går jämnt upp om resten är noll. För polynom uttrycks detta på följande sätt: Om <math>r(x)=0</math> så är <math>p(x)</math> delbart med <math>q(x)</math>, eller, <math>q(x)</math> är en ''delare'' till <math>p(x)</math>. Man skriver
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It is obvious that a division is exact if the remainder is zero. For polynomials this is expressed as follows: If <math>r(x)=0</math> then <math>p(x)</math> is divisible by <math>q(x)</math>, or, <math>q(x)</math> is a ''divisor'' of <math>p(x)</math>. One writes
{{Fristående formel||<math>\frac{p(x)}{q(x)} = k(x)\,\mbox{,}</math>}}
{{Fristående formel||<math>\frac{p(x)}{q(x)} = k(x)\,\mbox{,}</math>}}
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eller <math>\ p(x) = k(x)\cdot q(x)\,</math>.
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or <math>\ p(x) = k(x)\cdot q(x)\,</math>.
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== Polynomdivision ==
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== Polynomial long division ==
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Om <math>p(x)</math> är ett polynom med högre grad än polynomet <math>q(x)</math> så kan man dividera <math>p(x)</math> med <math>q(x)</math>. Det kan t.ex. göras genom att successivt subtrahera lämpliga multipler av <math>q(x)</math> från <math>p(x)</math> tills den återstående täljaren har lägre grad än <math>q(x)</math> i nämnaren.
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If <math>p(x)</math> is a polynomial of higher degree than polynomial <math>q(x)</math> then one can divide <math>p(x)</math> by <math>q(x)</math>. For example, this may be done by gradually subtracting appropriate multiples of <math>q(x)</math> from <math>p(x)</math> until a remaining numerator is of lower degree than the denominator <math>q(x)</math>.
<div class="exempel">
<div class="exempel">
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'''Exempel 2'''
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''' Example 2'''
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Utför polynomdivisionen <math>\ \frac{x^3 + x^2 -x +4}{x+2}\,</math>.
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Perform polynomial divisionen for <math>\ \frac{x^3 + x^2 -x +4}{x+2}\,</math>.
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Det första steget är att vi ''lägger till och drar ifrån'' en lämplig <math>x^2</math>-term i täljaren
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The first step is that we ''add and subtract'' an appropriate <math>x^2</math>-term in the numerator
{{Fristående formel||<math>\frac{x^3 + x^2 -x +4}{x+2} = \frac{x^3+2x^2-2x^2+x^2-x+4}{x+2}\,\mbox{.}</math>}}
{{Fristående formel||<math>\frac{x^3 + x^2 -x +4}{x+2} = \frac{x^3+2x^2-2x^2+x^2-x+4}{x+2}\,\mbox{.}</math>}}
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Anledningen till att vi gör detta är att nu kan deluttrycket <math>x^3+2x^2</math> skrivas som <math>x^2(x+2)</math> och förkortas med nämnaren
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The reason why we do this is now the sub-expression <math>x^3+2x^2</math> can be written as <math>x^2(x+2)</math> and cancellation with the denominator can be done,
{{Fristående formel||<math>\frac{x^2(x+2)-2x^2+x^2-x+4}{x+2} = x^2+\frac{-x^2-x+4}{x+2}\,\mbox{.}</math>}}
{{Fristående formel||<math>\frac{x^2(x+2)-2x^2+x^2-x+4}{x+2} = x^2+\frac{-x^2-x+4}{x+2}\,\mbox{.}</math>}}
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Sedan lägger vi till och drar ifrån en lämplig <math>x</math>-term så att den ledande <math>x^2</math>-termen i täljaren kan förkortas bort
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Then we add and subtract an appropriate <math>x</math>-term so that the leading <math>x^2</math>-term in the numerator can be cancelled,
{{Fristående formel||<math>\begin{align*} x^2+\frac{-x^2-2x+2x-x+4}{x+2} &= x^2+\frac{-x(x+2)+2x-x+4}{x+2}\\ &=x^2-x+\frac{x+4}{x+2}\,\mbox{.}\end{align*}</math>}}
{{Fristående formel||<math>\begin{align*} x^2+\frac{-x^2-2x+2x-x+4}{x+2} &= x^2+\frac{-x(x+2)+2x-x+4}{x+2}\\ &=x^2-x+\frac{x+4}{x+2}\,\mbox{.}\end{align*}</math>}}
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Det sista steget är att vi lägger till och drar ifrån en konstant
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The last step is that we add and subtract a constant
{{Fristående formel||<math>x^2-x+\frac{x+4}{x+2}=x^2-x+\frac{x+2-2+4}{x+2} = x^2-x+1+\frac{2}{x+2}\,\mbox{.}</math>}}
{{Fristående formel||<math>x^2-x+\frac{x+4}{x+2}=x^2-x+\frac{x+2-2+4}{x+2} = x^2-x+1+\frac{2}{x+2}\,\mbox{.}</math>}}
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Alltså gäller att
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Thus ending up with
{{Fristående formel||<math>\frac{x^3 + x^2 -x +4}{x+2} = x^2 -x + 1 + \frac{2}{x+2}\,\mbox{.}</math>}}
{{Fristående formel||<math>\frac{x^3 + x^2 -x +4}{x+2} = x^2 -x + 1 + \frac{2}{x+2}\,\mbox{.}</math>}}
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Kvoten är <math>x^2 -x + 1</math> och resten är <math>2</math>. Eftersom resten inte är noll går divisionen inte jämnt upp, dvs. <math>q(x)= x+2</math> är inte en ''delare'' till <math>p(x)=x^3 + x^2 -x +4</math>.
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The quotient is <math>x^2 -x + 1</math> and the remainder is <math>2</math>. Since the remainder is not zero division is not exact, that is, <math>q(x)= x+2</math>is not a ''divisor'' of <math>p(x)=x^3 + x^2 -x +4</math>.
</div>
</div>
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== Samband mellan faktorer och nollställen ==
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==The connection between factors and zeros ==
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Om <math>q(x)</math> är en delare till <math>p(x)</math> så gäller alltså att <math>p(x)=k(x)\cdot q(x)</math>. Vi har därmed ''faktoriserat'' <math>p(x)</math> . Man säger att <math>q(x)</math> är en faktor i <math>p(x)</math>. Speciellt gäller att om förstagradspolynomet <math>(x-a)</math> är en delare till <math>p(x)</math> så är <math>(x-a)</math> en faktor i <math>p(x)</math> , dvs.
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If <math>q(x)</math> is a divisor of <math>p(x)</math> then <math>p(x)=k(x)\cdot q(x)</math>. We have thus ''factorised'' <math>p(x)</math> . One says that <math>q(x)</math> is a factor of <math>p(x)</math>. Especially, if a polynomial of first degree <math>(x-a)</math> is a dividor of <math>p(x)</math> then <math>(x-a)</math> is a factor of <math>p(x)</math> , i.e.
{{Fristående formel||<math>p(x)= q(x)\cdot (x-a)\,\mbox{.}</math>}}
{{Fristående formel||<math>p(x)= q(x)\cdot (x-a)\,\mbox{.}</math>}}
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Eftersom <math>\ p(a)=q(a)\cdot (a-a)= q(a)\cdot 0 = 0\ </math> så betyder detta att <math>x=a</math> är ett nollställe till <math>p(x)</math>. Detta är precis innehållet i den s.k. ''faktorsatsen''.
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Since <math>\ p(a)=q(a)\cdot (a-a)= q(a)\cdot 0 = 0\ </math> this means that <math>x=a</math> is a zero of <math>p(x)</math>. This is exactly the content of the so-called ''factor theorem''.
<div class="regel">
<div class="regel">
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'''Faktorsatsen:'''
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'''Factor theorem:'''
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<math>(x-a)</math> är en delare till polynomet <math>p(x)</math> om och endast om <math>x=a</math> är ett nollställe till <math>p(x)</math>.
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<math>(x-a)</math> is a divisor of a polynomial <math>p(x)</math> if and only if <math>x=a</math> is a zero of <math>p(x)</math>.
</div>
</div>
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Observera att satsen gäller åt båda hållen, dvs. om man vet att <math>x=a</math> är ett nollställe till <math>p(x)</math> så vet man automatiskt att <math>p(x)</math> är delbart med <math>(x-a)</math>.
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Please note that the theorem applies in both directions, ie.. if we know that <math>x=a</math> is a zeros of <math>p(x)</math> we automatically would know that <math>p(x)</math> is divisible by <math>(x-a)</math>.
<div class="exempel">
<div class="exempel">
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'''Exempel 3'''
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''' Example 3'''
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Polynomet <math>p(x) = x^2-6x+8</math> kan faktoriseras som
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The polynomial <math>p(x) = x^2-6x+8</math> can be factorised as
{{Fristående formel||<math>x^2-6x+8 = (x-2)(x-4)</math>}}
{{Fristående formel||<math>x^2-6x+8 = (x-2)(x-4)</math>}}
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och har därför nollställena <math>x=2</math> och <math>x=4</math> (och inga andra). Det är precis dessa man får fram om man löser ekvationen <math>\ x^2-6x+8 = 0\,</math>.
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and has therefore zeros at <math>x=2</math> and <math>x=4</math> (and no others). It is precisely these that are obtained if one solves the equation <math>\ x^2-6x+8 = 0\,</math>.
</div>
</div>
<div class="exempel">
<div class="exempel">
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'''Exempel 4'''
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''' Example 4'''
<ol type="a">
<ol type="a">
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<li> Faktorisera polynomet <math>\ x^2-3x-10\,</math>.
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<li> Factorise the polynomial <math>\ x^2-3x-10\,</math>.
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By determining the polynomial zeros one automatically gets its factors according to the factor theorem. The quadratic equation <math>\ x^2-3x-10=0\ </math> has the solutions
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Genom att bestämma polynomets nollställen får man enligt faktorsatsen automatiskt dess faktorer. Andragradsekvationen <math>\ x^2-3x-10=0\ </math> har lösningarna
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{{Fristående formel||<math>x= \frac{3}{2} \pm \sqrt{\Bigl(\frac{3}{2}\Bigr)^2 - (-10)} = \frac{3}{2} \pm \frac{7}{2}\,\mbox{,}</math>}}
{{Fristående formel||<math>x= \frac{3}{2} \pm \sqrt{\Bigl(\frac{3}{2}\Bigr)^2 - (-10)} = \frac{3}{2} \pm \frac{7}{2}\,\mbox{,}</math>}}
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dvs. <math>x=-2</math> och <math>x=5</math>. Detta betyder att <math>\ x^2-3x-10=(x-(-2))(x-5)=(x+2)(x-5)\,</math>.
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i.e. <math>x=-2</math> and <math>x=5</math>. This means that <math>\ x^2-3x-10=(x-(-2))(x-5)=(x+2)(x-5)\,</math>.
</li>
</li>
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<li> Faktorisera polynomet <math>\ x^2+6x+9\,</math>.
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<li> Factorise the polynomial <math>\ x^2+6x+9\,</math>.
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Detta polynom har en dubbelrot
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This polynomial has a repeated root
{{Fristående formel||<math>x= -3 \pm \sqrt{\smash{(-3)^2 -9}\vphantom{i^2}} = -3</math>}}
{{Fristående formel||<math>x= -3 \pm \sqrt{\smash{(-3)^2 -9}\vphantom{i^2}} = -3</math>}}
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and thus <math>\ x^2+6x+9=(x-(-3))(x-(-3))=(x+3)^2\,</math>.
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och därmed är <math>\ x^2+6x+9=(x-(-3))(x-(-3))=(x+3)^2\,</math>.
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</li>
</li>
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<li> Faktorisera polynomet <math>\ x^2 -4x+5\,</math>.
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<li>Factorise the polynomial <math>\ x^2 -4x+5\,</math>.
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I detta fall har polynomet två komplexa rötter
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In this case, the polynomial has two complex roots
{{Fristående formel||<math>x= 2 \pm \sqrt{2^2 -5} = 2\pm \sqrt{-1} = 2\pm i</math>}}
{{Fristående formel||<math>x= 2 \pm \sqrt{2^2 -5} = 2\pm \sqrt{-1} = 2\pm i</math>}}
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och faktoriseringen blir <math>\ (x-(2-i))(x-(2+i))\,</math>.
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and when factorised will be <math>\ (x-(2-i))(x-(2+i))\,</math>.
</li>
</li>
Zeile 183: Zeile 180:
<div class="exempel">
<div class="exempel">
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'''Exempel 5'''
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''' Example 5'''
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Bestäm ett tredjegradspolynom med nollställena <math>1</math> , <math>-1</math> och <math>3</math>.
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Determine a cubic equation having zeros, <math>1</math> , <math>-1</math> and <math>3</math>.
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Polynomet måste enligt faktorsatsen ha faktorerna <math>(x-1)</math>, <math>(x+1)</math> och <math>(x-3)</math>. Multiplicerar vi ihop dessa faktorer får vi just ett tredjegradspolynom
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The polynomial according to the factor theorem, must have factors <math>(x-1)</math>, <math>(x+1)</math> and <math>(x-3)</math>. Multiplying these factors, we get a cubic equation
{{Fristående formel||<math>(x-1)(x+1)(x-3) = (x^2-1)(x-3)= x^3 -3x^2 -x+3\,\mbox{.}</math>}}
{{Fristående formel||<math>(x-1)(x+1)(x-3) = (x^2-1)(x-3)= x^3 -3x^2 -x+3\,\mbox{.}</math>}}
Zeile 196: Zeile 193:
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== Algebrans fundamentalsats ==
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== Fundamental theorem of algebra ==
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Vi införde i början av detta kapitel de komplexa talen för att kunna lösa andragradsekvationen <math>x^2=-1</math> och man kan nu ställa sig den lite mer teoretiska frågan om detta räcker, eller behöver vi uppfinna fler typer av tal för att kunna lösa andra mer komplicerade polynomekvationer. Svaret på den frågan är att det behöver vi inte göra utan det räcker med de komplexa talen. Den tyske matematikern Carl Friedrich Gauss bevisade år 1799 ''algebrans fundamentalsats'' som säger följande:
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We introduced at the beginning of this chapter, the complex numbers to enable us to solve quadratic equations <math>x^2=-1</math> and we can now ask ourselves the slightly more theoretical question, whether this is sufficient, or do we need to invent more types of numbers in order to solve other complicated polynomials? The answer to that question is that we need not as the complex numbers are enough. The German mathematician Carl Friedrich Gauss proved in the year 1799 the ''fundamental theorem of algebra'' which says the following:
<div class="regel">
<div class="regel">
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Varje polynom av grad <math>n\ge1</math> med komplexa koefficienter har minst ett nollställe bland de komplexa talen.
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Every polynomial of degree <math>n\ge1</math> with complex coefficients has at least one zero which is a complex number.
</div>
</div>
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Eftersom varje nollställe enligt faktorsatsen motsvaras av en faktor, kan man nu också fastställa följande sats:
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As every zero according to the the factor theorem is matched by a factor, we can now also state the following theorem:
<div class="regel">
<div class="regel">
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Varje polynom av grad <math>n\ge1</math> har exakt <math>n</math> stycken nollställen om varje nollställe räknas med sin ''multiplicitet''.
+
very polynomial of degree <math>n\ge1</math> has exactly <math>n</math> zeros if each zero is counted up to its multiplicity.
</div>
</div>
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(Med multiplicitet menas att ett dubbelt nollställe räknas 2 gånger, ett trippelnollställe 3 gånger, osv.)
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(By multiplicity is meant that a double zero is counted twice , a triple zero 3 times, etc.)
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Notera att dessa satser bara säger att det ''finns'' komplexa rötter till polynom men inte ''hur'' man räknar ut dessa. I allmänhet finns det ingen enkel metod att skriva upp en formel för rötterna, utan för polynomekvationer av högre gradtal får vi använda diverse knep för att lösa. Om vi håller oss till polynom med reella koefficienter så är ett av dessa knep som kan hjälpa oss att polynomets komplexa rötter alltid är komplexkonjugerade.
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Note that these theorems only say that there ''exists'' complex roots of polynomial, but not ''how'' to determine them. In general, there is no simple method to write a formula for the roots, but for polynomials of higher degree, we must use various devices to obtain a solution. If we restrict ourselves to polynomial with real coefficients, one of the devices that can help us is the knowledge that the complex roots of such polynomials always come in complex conjugate pairs .
<div class="exempel">
<div class="exempel">
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'''Exempel 6'''
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''' Example 6'''
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Visa att polynomet <math>p(x)=x^4-4x^3+6x^2-4x+5</math> har nollställena <math>x=i</math> och <math>x = 2-i</math>. Bestäm därefter övriga nollställen.
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Show that the polynomial <math>p(x)=x^4-4x^3+6x^2-4x+5</math> has zeros <math>x=i</math> and <math>x = 2-i</math>. Thus determine the other zeros.
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Vi har att
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We have
{{Fristående formel||<math>\begin{align*} p(i) &= i^4- 4i^3 +6i^2-4i+5 = 1+4i-6-4i+5=0\,\mbox{,}\\ p(2-i) &= (2-i)^4 -4(2-i)^3 + 6(2-i)^2 - 4(2-i) +5\,\mbox{.}\end{align*}</math>}}
{{Fristående formel||<math>\begin{align*} p(i) &= i^4- 4i^3 +6i^2-4i+5 = 1+4i-6-4i+5=0\,\mbox{,}\\ p(2-i) &= (2-i)^4 -4(2-i)^3 + 6(2-i)^2 - 4(2-i) +5\,\mbox{.}\end{align*}</math>}}
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För att räkna ut det sista uttrycket behöver vi bestämma
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In order to calculate the last term, we need to determine
{{Fristående formel||<math>\begin{align*} (2-i)^2 &= 4-4i+i^2 = 3-4i\,\mbox{,}\\ (2-i)^3 &= (3-4i)(2-i) = 6-3i-8i+4i^2 = 2-11i\,\mbox{,}\\ (2-i)^4 &= (2-11i)(2-i) = 4-2i-22i+11i^2= -7-24i\,\mbox{.}\end{align*}</math>}}
{{Fristående formel||<math>\begin{align*} (2-i)^2 &= 4-4i+i^2 = 3-4i\,\mbox{,}\\ (2-i)^3 &= (3-4i)(2-i) = 6-3i-8i+4i^2 = 2-11i\,\mbox{,}\\ (2-i)^4 &= (2-11i)(2-i) = 4-2i-22i+11i^2= -7-24i\,\mbox{.}\end{align*}</math>}}
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Detta ger att
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This gives that
{{Fristående formel||<math>\begin{align*} p(2-i) &= -7-24i-4(2-11i)+6(3-4i) -4(2-i) +5\\ &= -7-24i-8+44i+18-24i-8+4i+5=0\,\mbox{,}\end{align*}</math>}}
{{Fristående formel||<math>\begin{align*} p(2-i) &= -7-24i-4(2-11i)+6(3-4i) -4(2-i) +5\\ &= -7-24i-8+44i+18-24i-8+4i+5=0\,\mbox{,}\end{align*}</math>}}
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vilket visar att <math>i</math> och <math>2-i</math> är nollställen till polynomet.
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which proves that <math>i</math> and <math>2-i</math> are zeros of this polynomial.
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Eftersom polynomet har reella koefficienter kan vi direkt säga att de två andra nollställena är komplexkonjugatet av de två första nollställena, dvs. de två andra rötterna är <math>z=-i</math> och <math>z=2+i</math>.
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Since the polynomial has real coefficients, we can immediately say that the other two zeros are the complex conjugates of the first two zeros, i.e. the other two roots are <math>z=-i</math> and <math>z=2+i</math>.
</div>
</div>
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En konsekvens av algebrans fundamentalsats (och faktorsatsen) är att alla polynom kan faktoriseras i en produkt av komplexa förstagradsfaktorer. Detta gäller även polynom med reella koefficienter, men för dessa går det att multiplicera ihop förstagradsfaktorer som hör till komplexkonjugerade rötter och få en faktorisering helt med reella första- och andragradsfaktorer.
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One consequence of the fundamental theorem of algebra (and the factor theorem) is that all polynomials can be factored into a product of complex first order factors. This also applies to polynomials with real coefficients, but for such polynomials it is possible to multiply together the pair of factors belonging to complex conjugate roots. In this case the factorisation will consist of first and second order real factors.
<div class="exempel">
<div class="exempel">
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'''Exempel 7'''
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''' Example 7'''
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Visa att <math>x=1</math> är ett nollställe till <math>p(x)= x^3+x^2-2</math>. Faktorisera därefter <math>p(x)</math> i polynom med reella koefficienter, samt fullständigt i förstagradsfaktorer.
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Show that <math>x=1</math> is a zero of <math>p(x)= x^3+x^2-2</math>. Then first factorise <math>p(x)</math> into polynomials having real coefficients and then factorise <math>p(x)</math> completely into first order factors.
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Vi har att <math>\ p(1)= 1^3 + 1^2 -2 = 0\ </math> vilket visar att <math>x=1</math> är ett nollställe till polynomet. Enligt faktorsatsen betyder detta att <math>x-1</math> är en faktor i <math>p(x)</math>, dvs. att <math>p(x)</math> är delbar med <math>x-1</math>. Vi delar därför polynomet med <math>x-1</math> för att få återstående faktor om <math>x-1</math> bryts ut ur polynomet
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We have that <math>\ p(1)= 1^3 + 1^2 -2 = 0\ </math> which shows that <math>x=1</math> is a zero of the polynomial. According to the factor theorem, this means that <math>x-1</math> is a factor of <math>p(x)</math>, i.e. <math>p(x)</math> is divisible by <math>x-1</math>. We therefore divide the polynomial with <math>x-1</math> to get the remaining factors after <math>x-1</math> is factored out of the polynomial
{{Fristående formel||<math>\begin{align*} \frac{x^3+x^2-2}{x-1} &= \frac{x^2(x-1)+2x^2-2}{x-1} = x^2 + \frac{2x^2-2}{x-1} = x^2 + \frac{2x(x-1) +2x -2}{x-1}\\[4pt] &= x^2 + 2x + \frac{2x-2}{x-1} = x^2 + 2x + \frac{2(x-1)}{x-1} = x^2 + 2x + 2\,\mbox{.}\end{align*}</math>}}
{{Fristående formel||<math>\begin{align*} \frac{x^3+x^2-2}{x-1} &= \frac{x^2(x-1)+2x^2-2}{x-1} = x^2 + \frac{2x^2-2}{x-1} = x^2 + \frac{2x(x-1) +2x -2}{x-1}\\[4pt] &= x^2 + 2x + \frac{2x-2}{x-1} = x^2 + 2x + \frac{2(x-1)}{x-1} = x^2 + 2x + 2\,\mbox{.}\end{align*}</math>}}
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Alltså har vi att <math>\ p(x)= (x-1)(x^2+2x+2)\,</math>.
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So we have <math>\ p(x)= (x-1)(x^2+2x+2)\,</math> which is the first part of the problem.
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Nu återstår att faktorisera <math>x^2+2x+2</math>. Ekvationen <math>x^2+2x+2=0</math> har lösningarna
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It now remains to factorise <math>x^2+2x+2</math>. The equation <math>x^2+2x+2=0</math> has the solutions
{{Fristående formel||<math>x=-1\pm \sqrt{\smash{(-1)^2 -2}\vphantom{i^2}} = -1 \pm \sqrt{-1} = -1\pm i</math>}}
{{Fristående formel||<math>x=-1\pm \sqrt{\smash{(-1)^2 -2}\vphantom{i^2}} = -1 \pm \sqrt{-1} = -1\pm i</math>}}
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och därför har polynomet följande faktoriseringen i komplexa förstagradsfaktorer
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and therefore the polynomial has the following factorization into complex first order factors.
{{Fristående formel||<math>\begin{align*} x^3+x^2-2 = (x-1)(x^2+2x+2) &= (x-1)(x-(-1+i))(x-(-1-i))\\ &= (x-1)(x+1-i)(x+1+i)\,\mbox{.}\end{align*}</math>}}
{{Fristående formel||<math>\begin{align*} x^3+x^2-2 = (x-1)(x^2+2x+2) &= (x-1)(x-(-1+i))(x-(-1-i))\\ &= (x-1)(x+1-i)(x+1+i)\,\mbox{.}\end{align*}</math>}}
</div>
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Version vom 08:28, 26. Jul. 2008

 
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Content:

  • Factor theorem.
  • Polynomial long division
  • Fundamental theorem of algebra

{{Info| Learning outcomes:

After this section, you will have learned to:

  • Perform polynomial division.
  • Understand the relationship between factors and zeros of polynomials.
  • Know that a polynomial equation of degree n has n roots (including multiplicity).
  • Know that real polynomial equations have complex conjugate roots.

Polynom and equations

An expression of the form

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where \displaystyle n is an integer, is called a polynomial of degree \displaystyle n in an unknown variable \displaystyle x. The number \displaystyle a_1 is called the coefficient of \displaystyle x, \displaystyle a_2 the coefficient of \displaystyle x^2, etc. The constant \displaystyle a_0 is called the constant term.


Polynomials are essential for a large part of mathematics and have many properties which display great similarities with our integers, which means that we can do calculations with polynomials in a similar way as with integers.


Example 1


Compare the following integer written using a base 10,

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with the polynomial in \displaystyle x

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and then the following divisions,

  • \displaystyle \quad\frac{1353}{11} = 123 \qquad as \displaystyle \ 1353= 123\cdot 11\,,
  • \displaystyle \quad\frac{x^3 + 3x^2 + 5x + 3}{x+1} = x^2+2x+3\qquad since \displaystyle \ x^3 + 3x^2 + 5x + 3= (x^2+2x+3)(x+1)\,.

If \displaystyle p(x) is a polynomial of degree \displaystyle n then \displaystyle p(x)=0 is called a polynomial equation of degree \displaystyle n. If \displaystyle x=a is a number such that \displaystyle p(a)=0 then \displaystyle x=a is called a root, or solution to the equation. One also says that \displaystyle x=a is a zero of \displaystyle p(x).

As the above example showed polynomial, can be divided just like integers. Polynomial division, just like integer division usually is not exact. If, for example, \displaystyle 37 is divided by \displaystyle 5, one gets

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The calculation can also be written as \displaystyle \ 37= 7\cdot 5+2\,. The number 7 is called the quotient and the number 2 the remainder. One says that dividing 37 by 5 gives the quotient 7 and the remainder 2.


If \displaystyle p(x) and \displaystyle q(x) are polynomials one similarly can divide \displaystyle p(x) with \displaystyle q(x) and unambiguously determine polynomials \displaystyle k(x) and \displaystyle r(x) such that

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or \displaystyle \ p(x)= k(x)\cdot q(x)+r(x)\,. One here says that polynomial division has resulted in a quotient \displaystyle k(x) and remainder \displaystyle r(x).


It is obvious that a division is exact if the remainder is zero. For polynomials this is expressed as follows: If \displaystyle r(x)=0 then \displaystyle p(x) is divisible by \displaystyle q(x), or, \displaystyle q(x) is a divisor of \displaystyle p(x). One writes

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or \displaystyle \ p(x) = k(x)\cdot q(x)\,.


Polynomial long division

If \displaystyle p(x) is a polynomial of higher degree than polynomial \displaystyle q(x) then one can divide \displaystyle p(x) by \displaystyle q(x). For example, this may be done by gradually subtracting appropriate multiples of \displaystyle q(x) from \displaystyle p(x) until a remaining numerator is of lower degree than the denominator \displaystyle q(x).


Example 2


Perform polynomial divisionen for \displaystyle \ \frac{x^3 + x^2 -x +4}{x+2}\,.


The first step is that we add and subtract an appropriate \displaystyle x^2-term in the numerator

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The reason why we do this is now the sub-expression \displaystyle x^3+2x^2 can be written as \displaystyle x^2(x+2) and cancellation with the denominator can be done,

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Then we add and subtract an appropriate \displaystyle x-term so that the leading \displaystyle x^2-term in the numerator can be cancelled,

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The last step is that we add and subtract a constant

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Thus ending up with

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The quotient is \displaystyle x^2 -x + 1 and the remainder is \displaystyle 2. Since the remainder is not zero division is not exact, that is, \displaystyle q(x)= x+2is not a divisor of \displaystyle p(x)=x^3 + x^2 -x +4.


The connection between factors and zeros

If \displaystyle q(x) is a divisor of \displaystyle p(x) then \displaystyle p(x)=k(x)\cdot q(x). We have thus factorised \displaystyle p(x) . One says that \displaystyle q(x) is a factor of \displaystyle p(x). Especially, if a polynomial of first degree \displaystyle (x-a) is a dividor of \displaystyle p(x) then \displaystyle (x-a) is a factor of \displaystyle p(x) , i.e.

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Since \displaystyle \ p(a)=q(a)\cdot (a-a)= q(a)\cdot 0 = 0\ this means that \displaystyle x=a is a zero of \displaystyle p(x). This is exactly the content of the so-called factor theorem.

Factor theorem:

\displaystyle (x-a) is a divisor of a polynomial \displaystyle p(x) if and only if \displaystyle x=a is a zero of \displaystyle p(x).

Please note that the theorem applies in both directions, ie.. if we know that \displaystyle x=a is a zeros of \displaystyle p(x) we automatically would know that \displaystyle p(x) is divisible by \displaystyle (x-a).


Example 3


The polynomial \displaystyle p(x) = x^2-6x+8 can be factorised as

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and has therefore zeros at \displaystyle x=2 and \displaystyle x=4 (and no others). It is precisely these that are obtained if one solves the equation \displaystyle \ x^2-6x+8 = 0\,.

Example 4


  1. Factorise the polynomial \displaystyle \ x^2-3x-10\,. By determining the polynomial zeros one automatically gets its factors according to the factor theorem. The quadratic equation \displaystyle \ x^2-3x-10=0\ has the solutions
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    i.e. \displaystyle x=-2 and \displaystyle x=5. This means that \displaystyle \ x^2-3x-10=(x-(-2))(x-5)=(x+2)(x-5)\,.
  2. Factorise the polynomial \displaystyle \ x^2+6x+9\,. This polynomial has a repeated root
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    and thus \displaystyle \ x^2+6x+9=(x-(-3))(x-(-3))=(x+3)^2\,.
  3. Factorise the polynomial \displaystyle \ x^2 -4x+5\,. In this case, the polynomial has two complex roots
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    and when factorised will be \displaystyle \ (x-(2-i))(x-(2+i))\,.

Example 5


Determine a cubic equation having zeros, \displaystyle 1 , \displaystyle -1 and \displaystyle 3.


The polynomial according to the factor theorem, must have factors \displaystyle (x-1), \displaystyle (x+1) and \displaystyle (x-3). Multiplying these factors, we get a cubic equation

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Fundamental theorem of algebra

We introduced at the beginning of this chapter, the complex numbers to enable us to solve quadratic equations \displaystyle x^2=-1 and we can now ask ourselves the slightly more theoretical question, whether this is sufficient, or do we need to invent more types of numbers in order to solve other complicated polynomials? The answer to that question is that we need not as the complex numbers are enough. The German mathematician Carl Friedrich Gauss proved in the year 1799 the fundamental theorem of algebra which says the following:

Every polynomial of degree \displaystyle n\ge1 with complex coefficients has at least one zero which is a complex number.

As every zero according to the the factor theorem is matched by a factor, we can now also state the following theorem:

very polynomial of degree \displaystyle n\ge1 has exactly \displaystyle n zeros if each zero is counted up to its multiplicity.

(By multiplicity is meant that a double zero is counted twice , a triple zero 3 times, etc.)


Note that these theorems only say that there exists complex roots of polynomial, but not how to determine them. In general, there is no simple method to write a formula for the roots, but for polynomials of higher degree, we must use various devices to obtain a solution. If we restrict ourselves to polynomial with real coefficients, one of the devices that can help us is the knowledge that the complex roots of such polynomials always come in complex conjugate pairs .


Example 6


Show that the polynomial \displaystyle p(x)=x^4-4x^3+6x^2-4x+5 has zeros \displaystyle x=i and \displaystyle x = 2-i. Thus determine the other zeros.


We have

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In order to calculate the last term, we need to determine

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This gives that

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which proves that \displaystyle i and \displaystyle 2-i are zeros of this polynomial.


Since the polynomial has real coefficients, we can immediately say that the other two zeros are the complex conjugates of the first two zeros, i.e. the other two roots are \displaystyle z=-i and \displaystyle z=2+i.

One consequence of the fundamental theorem of algebra (and the factor theorem) is that all polynomials can be factored into a product of complex first order factors. This also applies to polynomials with real coefficients, but for such polynomials it is possible to multiply together the pair of factors belonging to complex conjugate roots. In this case the factorisation will consist of first and second order real factors.


Example 7


Show that \displaystyle x=1 is a zero of \displaystyle p(x)= x^3+x^2-2. Then first factorise \displaystyle p(x) into polynomials having real coefficients and then factorise \displaystyle p(x) completely into first order factors.


We have that \displaystyle \ p(1)= 1^3 + 1^2 -2 = 0\ which shows that \displaystyle x=1 is a zero of the polynomial. According to the factor theorem, this means that \displaystyle x-1 is a factor of \displaystyle p(x), i.e. \displaystyle p(x) is divisible by \displaystyle x-1. We therefore divide the polynomial with \displaystyle x-1 to get the remaining factors after \displaystyle x-1 is factored out of the polynomial

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So we have \displaystyle \ p(x)= (x-1)(x^2+2x+2)\, which is the first part of the problem.


It now remains to factorise \displaystyle x^2+2x+2. The equation \displaystyle x^2+2x+2=0 has the solutions

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and therefore the polynomial has the following factorization into complex first order factors.

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