Solution 3.4:5
From Förberedande kurs i matematik 2
m (Lösning 3.4:5 moved to Solution 3.4:5: Robot: moved page) |
m |
||
| (One intermediate revision not shown.) | |||
| Line 1: | Line 1: | ||
| - | {{ | + | A polynomial is said to have a triple root <math>z=c</math> if the equation contains the factor <math>(z-c)^3</math>. |
| - | < | + | |
| - | {{ | + | 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>}} | ||
| + | |||
| + | 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. | ||
| + | |||
| + | If we expand the right-hand side above, we get | ||
| + | |||
| + | {{Displayed math||<math>\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}</math>}} | ||
| + | |||
| + | 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>}} | ||
| + | |||
| + | 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>}} | ||
| + | |||
| + | 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>, | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | 3c(c-3c) &= -6\,,\\[5pt] | ||
| + | -6c^2 &= -6\,, | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | 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>a</math> and <math>b</math>, | ||
| + | |||
| + | |||
| + | <math>\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}</math> | ||
| + | |||
| + | |||
| + | Therefore, there are two different answers, | ||
| + | |||
| + | :*<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>. | ||
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) = 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.
