Lösung 2.1:3c

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The expression can be rewritten as <math> x^2+2\cdot 3\cdot x+3^2 </math> and then we see that it can be factorized, using the squaring rule <math> x^2+2ax+a^2=(x+a)^2</math>, as
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The expression can be rewritten as <math> x^2+2\cdot 3\cdot x+3^2 </math> and then we see that it can be factorized, using the squaring rule <math> x^2+2ax+a^2=(x+a)^2</math> as
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:<math> x^2+6x+9 =x^2+2\cdot 3\cdot x+3^2=(x+3)^2.</math>
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{{Displayed math||<math> x^2+6x+9 =x^2+2\cdot 3\cdot x+3^2=(x+3)^2\textrm{.}</math>}}
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Version vom 08:31, 23. Sep. 2008

The expression can be rewritten as \displaystyle x^2+2\cdot 3\cdot x+3^2 and then we see that it can be factorized, using the squaring rule \displaystyle x^2+2ax+a^2=(x+a)^2, as

Vorlage:Displayed math

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.1:3c“