From Förberedande kurs i matematik 2

Revision as of 14:07, 31 October 2008 (edit) Tek (Talk | contribs) m ← Previous diff		Current revision (18:11, 1 September 2011) (edit) (undo) Tek (Talk | contribs) (bug fix)
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	&= (z^2-2cz+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^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	+	&= 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>}}	+	{{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>}}
	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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	3c+d &= 0\,,\\[5pt]		3c+d &= 0\,,\\[5pt]
	3c(c+d) &= -6\,,\\[5pt]		3c(c+d) &= -6\,,\\[5pt]
-	-c^2(c-3d) &= a\,,\\[5pt]	+	-c^2(c+3d) &= a\,,\\[5pt]
	c^3d &= b\,\textrm{.}		c^3d &= b\,\textrm{.}
	\end{align}\right.</math>}}		\end{align}\right.</math>}}
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	<math>\begin{align}		<math>\begin{align}
-	c=1,\ d=-3:\quad a &= -1^2\cdot (1-3\cdot (-3)) = 8\,,\\[5pt]	+	c=1,\ d=-3:\quad a &= -1^2\cdot (1+3\cdot (-3)) = 8\,,\\[5pt]
	b &= 1^3\cdot (-3) = -3\,,\\[10pt]		b &= 1^3\cdot (-3) = -3\,,\\[10pt]
-	c=-1,\ d=3:\quad a &= -(-1)^2\cdot (-1-3\cdot 3) = 10\,,\\[5pt]	+	c=-1,\ d=3:\quad a &= -(-1)^2\cdot (-1+3\cdot 3) = -8\,,\\[5pt]
	b &= (-1)^3\cdot 3 = -3\,\textrm{.}		b &= (-1)^3\cdot 3 = -3\,\textrm{.}
	\end{align}</math>		\end{align}</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>,		:*<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>,
-	:*<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>.	+	:*<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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Current revision

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) = -8\,,\\[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=-8 and \displaystyle b=-3 give the triple root \displaystyle z=-1 and the single root \displaystyle z=3.