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>,
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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{{Abgesetzte Formel||<math>\begin{align}
(5x^3 + 3x^5)^2 &= (5x^3)^2 +2\cdot 5x^3\cdot 3x^5 +(3x^5)^{2} \\[3pt]
(5x^3 + 3x^5)^2 &= (5x^3)^2 +2\cdot 5x^3\cdot 3x^5 +(3x^5)^{2} \\[3pt]
&= 5^2x^{3\cdot 2} + 2\cdot 5\cdot 3\cdot x^{3+5}+ 3^2 x^{5\cdot 2}\\[3pt]
&= 5^2x^{3\cdot 2} + 2\cdot 5\cdot 3\cdot x^{3+5}+ 3^2 x^{5\cdot 2}\\[3pt]

Version vom 08:21, 22. Okt. 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 ,

\displaystyle \begin{align}

(5x^3 + 3x^5)^2 &= (5x^3)^2 +2\cdot 5x^3\cdot 3x^5 +(3x^5)^{2} \\[3pt] &= 5^2x^{3\cdot 2} + 2\cdot 5\cdot 3\cdot x^{3+5}+ 3^2 x^{5\cdot 2}\\[3pt] &= 25x^6 +30 x^8 +9x^{10}\\[3pt] &= 9x^{10} +30x^8 +25x^6\textrm{.} \end{align}

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.