Lösung 2.1:1h

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We expand the quadratic with the squaring rule <math> (a+b)^2=a^2+2ab+b^2 </math>, where <math> a=5x^3 </math> and <math> b=3x^5 </math>,
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We expand the quadratic with the squaring rule <math> (a+b)^2=a^2+2ab+b^2 </math>, where <math> a=5x^3 </math> and <math> b=3x^5 </math>
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{{Displayed math||<math>\begin{align}
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(5x^3 + 3x^5)^2 &= (5x^3)^2 +2\cdot 5x^3\cdot 3x^5 +(3x^5)^{2} \\[3pt]
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&= 5^2x^{3\cdot 2} + 2\cdot 5\cdot 3\cdot x^{3+5}+ 3^2 x^{5\cdot 2}\\[3pt]
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&= 25x^6 +30 x^8 +9x^{10}\\[3pt]
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&= 9x^{10} +30x^8 +25x^6\textrm{.}
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\end{align}</math>}}
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<math> \qquad
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Note: In the last line, we have moved the terms around so that the highest order term, <math> 9x^{10} </math>, comes first, followed by terms of decreasing order.
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\begin{align}
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(5x^3 + 3x^5)^2 &= (5x^3)^2 +2\cdot 5x^3\cdot 3x^5 +(3x^5)^{2} \\
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&= 5^2x^{3\cdot 2} + 2\cdot 5\cdot 3\cdot x^{3+5}+ 3^2 x^{5\cdot 2}\\
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&= 25x^6 +30 x^8 +9x^{10}\\
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&= 9x^{10} +30x^8 +25x^6
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\end{align}
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</math>
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NOTE: In the last line, we have moved the terms around so that the highest order term, <math> 9x^{10} </math>, comes first, followed by terms of decreasing order.
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Version vom 08:04, 23. Sep. 2008

We expand the quadratic with the squaring rule \displaystyle (a+b)^2=a^2+2ab+b^2 , where \displaystyle a=5x^3 and \displaystyle b=3x^5 ,

Vorlage:Displayed math

Note: In the last line, we have moved the terms around so that the highest order term, \displaystyle 9x^{10} , comes first, followed by terms of decreasing order.

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