Lösung 3.4:5
Aus Online Mathematik Brückenkurs 2
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| - | {{ | + | A polynomial is said to have a triple root |
| - | < | + | <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 | ||
| + | |||
| + | |||
| + | <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
z−c
3
For our equation, this means that the left-hand side can be factorized as
z−c3z−d
according to the factor theorem, where
We will now try to determine
If we expand the right-hand side above, we get
z−c3z−d=z−c2z−cz−d=
z2−2cz+c2
z−cz−d=z3−3cz2+3c2z−c3z−d=z4−3c+dz3+3cc+dz2−c2c−3dz+c3d
and this means that we must have
3c+dz3+3cc+dz2−c2c−3dz+c3d
Because two polynomials are equal if an only if their coefficients are equal, this gives
3c+d=03cc+d=−6−c2c−3d=ac3d=b
From the first equation, we obtain
c−3c=−6−6c2=−6
i.e.
\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}.
