Aus Online Mathematik Brückenkurs 2

Version vom 15:13, 16. Sep. 2008 (bearbeiten) Tekbot (Diskussion | Beiträge) K (Lösning 3.4:5 moved to Solution 3.4:5: Robot: moved page) ← Zum vorherigen Versionsunterschied		Version vom 11:09, 28. Okt. 2008 (bearbeiten) (rückgängig) Ian (Diskussion | Beiträge) Zum nächsten Versionsunterschied →
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-	{{NAVCONTENT_START}}	+	A polynomial is said to have a triple root
-	<center> [[Image:3_4_5.gif]] </center>	+	<math>z=c</math>
-	{{NAVCONTENT_STOP}}	+	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
		+
		+
		+	<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\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
		+
		+
		+	<math>\begin{align}
		+	& \left( z-c \right)^{3}\left( z-d \right)=\left( z-c \right)^{2}\left( z-c \right)\left( z-d \right) \\
		+	& =\left( z^{2}-2cz+c^{2} \right)\left( z-c \right)\left( z-d \right) \\
		+	& =\left( z^{3}-3cz^{2}+3c^{2}z-c^{3} \right)\left( z-d \right) \\
		+	& =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 \\
		+	\end{align}</math>
		+
		+
		+	and this means that we must have
		+
		+
		+	<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
		+
		+
		+	<math>\left\{ \begin{array}{*{35}l}
		+	3c+d=0 \\
		+	3c\left( c+d \right)=-6 \\
		+	-c^{2}\left( c-3d \right)=a \\
		+	c^{3}d=b \\
		+	\end{array} \right.</math>
		+
		+
		+	From the first equation, we obtain
		+	<math>d=-\text{3}c\text{ }</math>
		+	and substituting this into the second equation gives us an equation for
		+	<math>c</math>,
		+
		+
		+	<math>\begin{align}
		+	& 3c\left( c-3c \right)=-6 \\
		+	& -6c^{2}=-6 \\
		+	\end{align}</math>
		+
		+
		+	i.e.
		+	<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>
		+	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>
		+	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 11:09, 28. Okt. 2008

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

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

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

according to the factor theorem, where \displaystyle z=c is the triple root and \displaystyle z=d\text{ } 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} & \left( z-c \right)^{3}\left( z-d \right)=\left( z-c \right)^{2}\left( z-c \right)\left( z-d \right) \\ & =\left( z^{2}-2cz+c^{2} \right)\left( z-c \right)\left( z-d \right) \\ & =\left( z^{3}-3cz^{2}+3c^{2}z-c^{3} \right)\left( z-d \right) \\ & =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 \\ \end{align}

and this means that we must have

\displaystyle 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.

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

\displaystyle \left\{ \begin{array}{*{35}l} 3c+d=0 \\ 3c\left( c+d \right)=-6 \\ -c^{2}\left( c-3d \right)=a \\ c^{3}d=b \\ \end{array} \right.

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

\displaystyle \begin{align} & 3c\left( c-3c \right)=-6 \\ & -6c^{2}=-6 \\ \end{align}

i.e. \displaystyle c=-\text{1 } or \displaystyle c=\text{1}. The relation \displaystyle d=-\text{3}c\text{ } 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}\centerdot \left( 1-3\centerdot \left( -3 \right) \right)=8 \\ & b=1^{3}\centerdot \left( -3 \right)=-3 \\ \end{align}

\displaystyle \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}

Therefore, there are two different answers:

\displaystyle \bullet \quad a=\text{8 } and \displaystyle b=-\text{3} give the triple root \displaystyle z=\text{1} and the single root \displaystyle z=-\text{3};

\displaystyle \bullet \quad a=10 and \displaystyle b=-\text{3 }give the triple root \displaystyle z=-\text{1 } and the single root \displaystyle z=\text{3}.