Lösung 2.1:2d

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K (Lösning 2.1:2d moved to Solution 2.1:2d: Robot: moved page)
K
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The three brackets can be multiplied together in any order, but just in this case it seems appropriate to start by multiplying the first two brackets because we can carry out the multiplication directly by using the conjugate rule,
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<!--center> [[Image:2_1_2d.gif]] </center-->
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The three brackets can be multiplied together in any order, but just in this case it seems appropriate to start by multiplying the first two brackets because we can carry out the multiplication directly by using the conjugate rule.
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<math>
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{{Displayed math||<math>\begin{align}
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\qquad
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\begin{align}
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(3x^2+2)(3x^2-2)(9x^4+4) &= \big( (3x^2)^2-2^2 \big) (9x^4+4)\\
(3x^2+2)(3x^2-2)(9x^4+4) &= \big( (3x^2)^2-2^2 \big) (9x^4+4)\\
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&= (9x^4-4)(9x^4+4)
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&= (9x^4-4)(9x^4+4)\,\textrm{.}
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\end{align}
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\end{align}</math>}}
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</math>
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We happend to obtain an expression which can also be multiplied using the conjugate rule.
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We happend to obtain an expression which can also be multiplied using the conjugate rule,
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<math>
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{{Displayed math||<math>(9x^4-4)(9x^4+4)=(9x^4)^2-4^2=81x^8-16\,\textrm{.}</math>}}
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\qquad
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\begin{align}
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(9x^4-4)(9x^4+4)=(9x^4)^2-4^2=81x^8-16
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\end{align}
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</math>
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NOTE: If we had instead multiplied together the terms in the first and third brackets first, we would also have got the right answer, but the calculations would have been lengthier.
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Note: If we had instead multiplied together the terms in the first and third brackets first, we would also have got the right answer, but the calculations would have been lengthier.
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Version vom 08:22, 23. Sep. 2008

The three brackets can be multiplied together in any order, but just in this case it seems appropriate to start by multiplying the first two brackets because we can carry out the multiplication directly by using the conjugate rule,

Vorlage:Displayed math

We happend to obtain an expression which can also be multiplied using the conjugate rule,

Vorlage:Displayed math

Note: If we had instead multiplied together the terms in the first and third brackets first, we would also have got the right answer, but the calculations would have been lengthier.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.1:2d“