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Lösung 3.4:5

Aus Online Mathematik Brückenkurs 2

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A polynomial is said to have a triple root
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A polynomial is said to have a triple root <math>z=c</math> if the equation contains the factor <math>(z-c)^3</math>.
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<math>z=c</math>
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if the equation contains the factor
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<math>\left( z-c \right)^{3}</math>
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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
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{{Displayed math||<math>z^4-6z^2+az+b = (z-c)^3(z-d)</math>}}
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<math>z^{4}-6z^{2}+az+b=\left( z-c \right)^{3}\left( z-d \right)</math>
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according to the factor theorem, where <math>z=c</math> is the triple root and
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<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).
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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.
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according to the factor theorem, where
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<math>z=c</math>
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is the triple root and
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<math>z=d\text{ }</math>
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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).
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We will now try to determine
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<math>a</math>,
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<math>b</math>,
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<math>c</math>
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and
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<math>d</math>
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so that both sides in the factorization above agree.
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If we expand the right-hand side above, we get
If we expand the right-hand side above, we get
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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(z-c)^3(z-d)
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& \left( z-c \right)^{3}\left( z-d \right)=\left( z-c \right)^{2}\left( z-c \right)\left( z-d \right) \\
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&= (z-c)^2(z-c)(z-d)\\[5pt]
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& =\left( z^{2}-2cz+c^{2} \right)\left( z-c \right)\left( z-d \right) \\
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&= (z^2-2cz+c^2)(z-c)(z-d)\\[5pt]
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& =\left( z^{3}-3cz^{2}+3c^{2}z-c^{3} \right)\left( z-d \right) \\
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&= (z^3-3cz^2+3c^2z-c^3)(z-d)\\[5pt]
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& =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 \\
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&= z^4-(3c+d)z^3+3c(c+d)z^2-c^2(c-3d)z+c^3d
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\end{align}</math>
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\end{align}</math>}}
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and this means that we must have
and this means that we must have
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{{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>}}
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<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>
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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
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{{Displayed math||<math>\left\{\begin{align}
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3c+d &= 0\,,\\[5pt]
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3c(c+d) &= -6\,,\\[5pt]
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-c^2(c-3d) &= a\,,\\[5pt]
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c^3d &= b\,\textrm{.}
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\end{align}\right.</math>}}
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<math>\left\{ \begin{array}{*{35}l}
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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>,
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3c+d=0 \\
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3c\left( c+d \right)=-6 \\
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-c^{2}\left( c-3d \right)=a \\
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c^{3}d=b \\
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\end{array} \right.</math>
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{{Displayed math||<math>\begin{align}
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3c(c-3c) &= -6\,,\\[5pt]
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-6c^2 &= -6\,,
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\end{align}</math>}}
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From the first equation, we obtain
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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
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<math>d=-\text{3}c\text{ }</math>
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<math>a</math> and <math>b</math>,
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and substituting this into the second equation gives us an equation for
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<math>c</math>,
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<math>\begin{align}
<math>\begin{align}
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& 3c\left( c-3c \right)=-6 \\
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c=1,\ d=-3:\quad a &= -1^2\cdot (1-3\cdot (-3)) = 8\,,\\[5pt]
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& -6c^{2}=-6 \\
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b &= 1^3\cdot (-3) = -3\,,\\[10pt]
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c=-1,\ d=3:\quad a &= -(-1)^2\cdot (-1-3\cdot 3) = 10\,,\\[5pt]
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b &= (-1)^3\cdot 3 = -3\,\textrm{.}
\end{align}</math>
\end{align}</math>
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i.e.
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Therefore, there are two different answers,
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<math>c=-\text{1 }</math>
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or
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<math>c=\text{1}</math>. The relation
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<math>d=-\text{3}c\text{ }</math>
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gives that the corresponding values for
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<math>d</math>
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are
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<math>d=3</math>
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and
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<math>d=-3</math>. The two last equations give us the corresponding values for
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<math>a</math>
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and
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<math>b</math>,
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<math>\begin{align}
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& c=1,\ d=-3:\quad a=-1^{2}\centerdot \left( 1-3\centerdot \left( -3 \right) \right)=8 \\
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& b=1^{3}\centerdot \left( -3 \right)=-3 \\
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\end{align}</math>
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<math>\begin{align}
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& c=-1,\ d=3:\quad a=-\left( -1 \right)^{2}\centerdot \left( -1-3\centerdot 3 \right)=10 \\
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& b=\left( -1 \right)^{3}\centerdot 3=-3 \\
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\end{align}</math>
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Therefore, there are two different answers:
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<math>\bullet \quad a=\text{8 }</math>
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:*<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>,
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and
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<math>b=-\text{3}</math>
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give the triple root
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<math>z=\text{1}</math>
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and the single root
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<math>z=-\text{3}</math>;
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<math>\bullet \quad a=10</math>
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:*<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>.
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and
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<math>b=-\text{3 }</math>give the triple root
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<math>z=-\text{1 }</math>
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and the single root
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<math>z=\text{3}</math>.
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Version vom 14:07, 31. Okt. 2008

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

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

z4−6z2+az+b=(z−c)3(z−d)

according to the factor theorem, where z=c is the triple root and 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 a, b, c and d so that both sides in the factorization above agree.

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

(z−c)3(z−d)=(z−c)2(z−c)(z−d)=(z2−2cz+c2)(z−c)(z−d)=(z3−3cz2+3c2z−c3)(z−d)=z4−(3c+d)z3+3c(c+d)z2−c2(c−3d)z+c3d

and this means that we must have

z4−6z2+az+b=z4−(3c+d)z3+3c(c+d)z2−c2(c−3d)z+c3d.

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

3c+d3c(c+d)−c2(c−3d)c3d=0=−6=a=b.

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

3c(c−3c)−6c2=−6=−6

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


c=1 d=−3:abc=−1 d=3:ab=−12(1−3(−3))=8=13(−3)=−3=−(−1)2(−1−33)=10=(−1)33=−3.


Therefore, there are two different answers,

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