Lösung 3.4:2
Aus Online Mathematik Brückenkurs 2
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If the equation has the root <math>z=1</math>, this means, according to the factor rule, that the equation must contain the factor <math>z-1</math>, i.e. the polynomial on the left-hand side can be written as | If the equation has the root <math>z=1</math>, this means, according to the factor rule, that the equation must contain the factor <math>z-1</math>, i.e. the polynomial on the left-hand side can be written as | ||
| - | {{ | + | {{Abgesetzte Formel||<math>z^3-3z^2+4z-2 = (z^2+Az+B)(z-1)</math>}} |
for some constants <math>A</math> and <math>B</math>. We can determine the second unknown factor using polynomial division, | for some constants <math>A</math> and <math>B</math>. We can determine the second unknown factor using polynomial division, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
z^2+Az+B | z^2+Az+B | ||
&= \frac{z^3-3z^2+4z-2}{z-1}\\[5pt] | &= \frac{z^3-3z^2+4z-2}{z-1}\\[5pt] | ||
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Thus, the equation can be written as | Thus, the equation can be written as | ||
| - | {{ | + | {{Abgesetzte Formel||<math>(z-1)(z^2-2z+2) = 0\,\textrm{.}</math>}} |
The advantage of writing the equation in this factorized form is that we can now conclude that the equation's two other roots must be zeros of the factor | The advantage of writing the equation in this factorized form is that we can now conclude that the equation's two other roots must be zeros of the factor | ||
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Hence, we determine the roots by solving the equation | Hence, we determine the roots by solving the equation | ||
| - | {{ | + | {{Abgesetzte Formel||<math>z^2-2z+2 = 0\,\textrm{.}</math>}} |
Completing the square gives | Completing the square gives | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
(z-1)^2-1^2+2 &= 0\,,\\[5pt] | (z-1)^2-1^2+2 &= 0\,,\\[5pt] | ||
(z-1)^2 &= -1\,, | (z-1)^2 &= -1\,, | ||
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Note: Writing | Note: Writing | ||
| - | {{ | + | {{Abgesetzte Formel||<math>z^3-3z^2+4z-2 = \bigl((z-3)z+4\bigr)z-2</math>}} |
is known as the Horner scheme and is used to reduce the amount of the arithmetical work. | is known as the Horner scheme and is used to reduce the amount of the arithmetical work. | ||
Version vom 13:15, 10. Mär. 2009
If the equation has the root \displaystyle z=1, this means, according to the factor rule, that the equation must contain the factor \displaystyle z-1, i.e. the polynomial on the left-hand side can be written as
| \displaystyle z^3-3z^2+4z-2 = (z^2+Az+B)(z-1) |
for some constants \displaystyle A and \displaystyle B. We can determine the second unknown factor using polynomial division,
| \displaystyle \begin{align}
z^2+Az+B &= \frac{z^3-3z^2+4z-2}{z-1}\\[5pt] &= \frac{z^3-z^2+z^2-3z^2+4z-2}{z-1}\\[5pt] &= \frac{z^2(z-1)-2z^2+4z-2}{z-1}\\[5pt] &= z^2 + \frac{-2z^2+4z-2}{z-1}\\[5pt] &= z^2 + \frac{-2z^2+2z-2z+4z-2}{z-1}\\[5pt] &= z^2 + \frac{-2z(z-1)+2z-2}{z-1}\\[5pt] &= z^2 - 2z + \frac{2z-2}{z-1}\\[5pt] &= z^2 - 2z + \frac{2(z-1)}{z-1}\\[5pt] &= z^2 - 2z + 2\,\textrm{.} \end{align} |
Thus, the equation can be written as
| \displaystyle (z-1)(z^2-2z+2) = 0\,\textrm{.} |
The advantage of writing the equation in this factorized form is that we can now conclude that the equation's two other roots must be zeros of the factor \displaystyle z^2-2z+2. This is because the left-hand side is zero only when at least one of the factors \displaystyle z-1 or \displaystyle z^2-2z+2 is zero, and we see directly that \displaystyle z-1 is zero only when \displaystyle z=1\,.
Hence, we determine the roots by solving the equation
| \displaystyle z^2-2z+2 = 0\,\textrm{.} |
Completing the square gives
| \displaystyle \begin{align}
(z-1)^2-1^2+2 &= 0\,,\\[5pt] (z-1)^2 &= -1\,, \end{align} |
and taking the root gives that \displaystyle z-1=\pm i, i.e. \displaystyle z=1-i and \displaystyle z=1+i\,.
The equation's other roots are \displaystyle z=1-i and \displaystyle z=1+i\,.
As an extra check, we investigate whether \displaystyle z = 1 \pm i really are roots of the equation.
\displaystyle \begin{align} z = 1+i:\quad z^3-3z^2+4z-2 &= \bigl((z-3)z+4\bigr)z-2\\[5pt] &= \bigl((1+i-3)(1+i)+4\bigr)(1+i)-2\\[5pt] &= \bigl((-2+i)(1+i)+4\bigr)(1+i)-2\\[5pt] &= (-2+i-2i-1+4)(1+i)-2\\[5pt] &= (1-i)(1+i)-2\\[5pt] &= 1^2 - i^2 - 2\\[5pt] &= 1+1-2\\[5pt] &= 0\,,\\[10pt] z = 1-i:\quad z^3-3z^2+4z-2 &= \bigl((z-3)z+4\bigr)z-2\\[5pt] &= \bigl((1-i-3)(1-i)+4\bigr)(1-i)-2\\[5pt] &= \bigl((-2-i)(1-i)+4\bigr)(1-i)-2\\[5pt] &= (-2-i+2i-1+4)(1-i)-2\\[5pt] &= (1+i)(1-i)-2\\[5pt] &= 1^2-i^2-2\\[5pt] &= 1+1-2\\[5pt] &= 0\,\textrm{.} \end{align}
Note: Writing
| \displaystyle z^3-3z^2+4z-2 = \bigl((z-3)z+4\bigr)z-2 |
is known as the Horner scheme and is used to reduce the amount of the arithmetical work.
