Lösung 3.4:5

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
Zeile 1: Zeile 1:
-
A polynomial is said to have a triple root
+
A polynomial is said to have a triple root <math>z=c</math> if the equation contains the factor <math>(z-c)^3</math>.
-
<math>z=c</math>
+
-
if the equation contains the factor
+
-
<math>\left( z-c \right)^{3}</math>
+
-
 
+
For our equation, this means that the left-hand side can be factorized as
For our equation, this means that the left-hand side can be factorized as
 +
{{Displayed math||<math>z^4-6z^2+az+b = (z-c)^3(z-d)</math>}}
-
<math>z^{4}-6z^{2}+az+b=\left( z-c \right)^{3}\left( z-d \right)</math>
+
according to the factor theorem, where <math>z=c</math> is the triple root and
 +
<math>z=d</math> is the equation's fourth root (according to the fundamental theorem of algebra, a fourth-order equation always has four roots, taking into account multiplicity).
-
 
+
We will now try to determine <math>a</math>, <math>b</math>, <math>c</math> and <math>d</math> so that both sides in the factorization above agree.
-
according to the factor theorem, where
+
-
<math>z=c</math>
+
-
is the triple root and
+
-
<math>z=d\text{ }</math>
+
-
is the equation's fourth root (according to the fundamental theorem of algebra, a fourth-order equation always has four roots, taking into account multiplicity).
+
-
 
+
-
We will now try to determine
+
-
<math>a</math>,
+
-
<math>b</math>,
+
-
<math>c</math>
+
-
and
+
-
<math>d</math>
+
-
so that both sides in the factorization above agree.
+
If we expand the right-hand side above, we get
If we expand the right-hand side above, we get
-
 
+
{{Displayed math||<math>\begin{align}
-
<math>\begin{align}
+
(z-c)^3(z-d)
-
& \left( z-c \right)^{3}\left( z-d \right)=\left( z-c \right)^{2}\left( z-c \right)\left( z-d \right) \\
+
&= (z-c)^2(z-c)(z-d)\\[5pt]
-
& =\left( z^{2}-2cz+c^{2} \right)\left( z-c \right)\left( z-d \right) \\
+
&= (z^2-2cz+c^2)(z-c)(z-d)\\[5pt]
-
& =\left( z^{3}-3cz^{2}+3c^{2}z-c^{3} \right)\left( z-d \right) \\
+
&= (z^3-3cz^2+3c^2z-c^3)(z-d)\\[5pt]
-
& =z^{4}-\left( 3c+d \right)z^{3}+3c\left( c+d \right)z^{2}-c^{2}\left( c-3d \right)z+c^{3}d \\
+
&= z^4-(3c+d)z^3+3c(c+d)z^2-c^2(c-3d)z+c^3d
-
\end{align}</math>
+
\end{align}</math>}}
-
 
+
and this means that we must have
and this means that we must have
-
 
+
{{Displayed math||<math>z^4-6z^2+az+b = z^4-(3c+d)z^3+3c(c+d)z^2-c^2(c-3d)z+c^3d\,\textrm{.}</math>}}
-
<math>z^{4}-6z^{2}+az+b=z^{4}-\left( 3c+d \right)z^{3}+3c\left( c+d \right)z^{2}-c^{2}\left( c-3d \right)z+c^{3}d.</math>
+
-
 
+
Because two polynomials are equal if an only if their coefficients are equal, this gives
Because two polynomials are equal if an only if their coefficients are equal, this gives
 +
{{Displayed math||<math>\left\{\begin{align}
 +
3c+d &= 0\,,\\[5pt]
 +
3c(c+d) &= -6\,,\\[5pt]
 +
-c^2(c-3d) &= a\,,\\[5pt]
 +
c^3d &= b\,\textrm{.}
 +
\end{align}\right.</math>}}
-
<math>\left\{ \begin{array}{*{35}l}
+
From the first equation, we obtain <math>d=-3c</math> and substituting this into the second equation gives us an equation for <math>c</math>,
-
3c+d=0 \\
+
-
3c\left( c+d \right)=-6 \\
+
-
-c^{2}\left( c-3d \right)=a \\
+
-
c^{3}d=b \\
+
-
\end{array} \right.</math>
+
 +
{{Displayed math||<math>\begin{align}
 +
3c(c-3c) &= -6\,,\\[5pt]
 +
-6c^2 &= -6\,,
 +
\end{align}</math>}}
-
From the first equation, we obtain
+
i.e. <math>c=-1</math> or <math>c=1</math>. The relation <math>d=-3c</math> gives that the corresponding values for <math>d</math> are <math>d=3</math> and <math>d=-3</math>. The two last equations give us the corresponding values for
-
<math>d=-\text{3}c\text{ }</math>
+
<math>a</math> and <math>b</math>,
-
and substituting this into the second equation gives us an equation for
+
-
<math>c</math>,
+
<math>\begin{align}
<math>\begin{align}
-
& 3c\left( c-3c \right)=-6 \\
+
c=1,\ d=-3:\quad a &= -1^2\cdot (1-3\cdot (-3)) = 8\,,\\[5pt]
-
& -6c^{2}=-6 \\
+
b &= 1^3\cdot (-3) = -3\,,\\[10pt]
 +
c=-1,\ d=3:\quad a &= -(-1)^2\cdot (-1-3\cdot 3) = 10\,,\\[5pt]
 +
b &= (-1)^3\cdot 3 = -3\,\textrm{.}
\end{align}</math>
\end{align}</math>
-
i.e.
+
Therefore, there are two different answers,
-
<math>c=-\text{1 }</math>
+
-
or
+
-
<math>c=\text{1}</math>. The relation
+
-
<math>d=-\text{3}c\text{ }</math>
+
-
gives that the corresponding values for
+
-
<math>d</math>
+
-
are
+
-
<math>d=3</math>
+
-
and
+
-
<math>d=-3</math>. The two last equations give us the corresponding values for
+
-
<math>a</math>
+
-
and
+
-
<math>b</math>,
+
-
 
+
-
 
+
-
<math>\begin{align}
+
-
& c=1,\ d=-3:\quad a=-1^{2}\centerdot \left( 1-3\centerdot \left( -3 \right) \right)=8 \\
+
-
& b=1^{3}\centerdot \left( -3 \right)=-3 \\
+
-
\end{align}</math>
+
-
 
+
-
 
+
-
 
+
-
<math>\begin{align}
+
-
& c=-1,\ d=3:\quad a=-\left( -1 \right)^{2}\centerdot \left( -1-3\centerdot 3 \right)=10 \\
+
-
& b=\left( -1 \right)^{3}\centerdot 3=-3 \\
+
-
\end{align}</math>
+
-
 
+
-
 
+
-
Therefore, there are two different answers:
+
-
 
+
-
<math>\bullet \quad a=\text{8 }</math>
+
:*<math>a=8</math> and <math>b=-3</math> give the triple root <math>z=1</math> and the single root <math>z=-3</math>,
-
and
+
-
<math>b=-\text{3}</math>
+
-
give the triple root
+
-
<math>z=\text{1}</math>
+
-
and the single root
+
-
<math>z=-\text{3}</math>;
+
-
<math>\bullet \quad a=10</math>
+
:*<math>a=10</math> and <math>b=-3</math> give the triple root <math>z=-1</math> and the single root <math>z=3</math>.
-
and
+
-
<math>b=-\text{3 }</math>give the triple root
+
-
<math>z=-\text{1 }</math>
+
-
and the single root
+
-
<math>z=\text{3}</math>.
+

Version vom 14:07, 31. Okt. 2008

A polynomial is said to have a triple root \displaystyle z=c if the equation contains the factor \displaystyle (z-c)^3.

For our equation, this means that the left-hand side can be factorized as

\displaystyle z^4-6z^2+az+b = (z-c)^3(z-d)

according to the factor theorem, where \displaystyle z=c is the triple root and \displaystyle z=d is the equation's fourth root (according to the fundamental theorem of algebra, a fourth-order equation always has four roots, taking into account multiplicity).

We will now try to determine \displaystyle a, \displaystyle b, \displaystyle c and \displaystyle d so that both sides in the factorization above agree.

If we expand the right-hand side above, we get

\displaystyle \begin{align}

(z-c)^3(z-d) &= (z-c)^2(z-c)(z-d)\\[5pt] &= (z^2-2cz+c^2)(z-c)(z-d)\\[5pt] &= (z^3-3cz^2+3c^2z-c^3)(z-d)\\[5pt] &= z^4-(3c+d)z^3+3c(c+d)z^2-c^2(c-3d)z+c^3d \end{align}

and this means that we must have

\displaystyle z^4-6z^2+az+b = z^4-(3c+d)z^3+3c(c+d)z^2-c^2(c-3d)z+c^3d\,\textrm{.}

Because two polynomials are equal if an only if their coefficients are equal, this gives

\displaystyle \left\{\begin{align}

3c+d &= 0\,,\\[5pt] 3c(c+d) &= -6\,,\\[5pt] -c^2(c-3d) &= a\,,\\[5pt] c^3d &= b\,\textrm{.} \end{align}\right.

From the first equation, we obtain \displaystyle d=-3c and substituting this into the second equation gives us an equation for \displaystyle c,

\displaystyle \begin{align}

3c(c-3c) &= -6\,,\\[5pt] -6c^2 &= -6\,, \end{align}

i.e. \displaystyle c=-1 or \displaystyle c=1. The relation \displaystyle d=-3c gives that the corresponding values for \displaystyle d are \displaystyle d=3 and \displaystyle d=-3. The two last equations give us the corresponding values for \displaystyle a and \displaystyle b,


\displaystyle \begin{align} c=1,\ d=-3:\quad a &= -1^2\cdot (1-3\cdot (-3)) = 8\,,\\[5pt] b &= 1^3\cdot (-3) = -3\,,\\[10pt] c=-1,\ d=3:\quad a &= -(-1)^2\cdot (-1-3\cdot 3) = 10\,,\\[5pt] b &= (-1)^3\cdot 3 = -3\,\textrm{.} \end{align}


Therefore, there are two different answers,

  • \displaystyle a=8 and \displaystyle b=-3 give the triple root \displaystyle z=1 and the single root \displaystyle z=-3,
  • \displaystyle a=10 and \displaystyle b=-3 give the triple root \displaystyle z=-1 and the single root \displaystyle z=3.
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.4:5“