Lösung 2.1:5c

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The fraction can be further simplified if it is possible to factorize and eliminate common factors
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<center> [[Image:2_1_5c.gif]] </center>
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from the numerator and denominator. Both numerator and denominator are already factorized to a certain extent, but we can go further with the numerator and break it up into linear factors by using the conjugate rule:
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<math>\begin{align}
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& 3x^{2}-12=3\left( x^{2}-4 \right)=3\left( x+2 \right)\left( x-2 \right) \\
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& \\
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& x^{2}-1=\left( x+1 \right)\left( x-1 \right) \\
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\end{align}</math>
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The whole expression is therefore equal to
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<math>\frac{3\left( x+2 \right)\left( x-2 \right)\left( x+1 \right)\left( x-1 \right)}{\left( x+1 \right)\left( x+2 \right)}=3\left( x-2 \right)\left( x-1 \right)</math>
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NOTE: One can of course expand out the expression to get
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<math>3x^{2}-9x+6</math>
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as the answer.

Version vom 09:26, 16. Sep. 2008

The fraction can be further simplified if it is possible to factorize and eliminate common factors from the numerator and denominator. Both numerator and denominator are already factorized to a certain extent, but we can go further with the numerator and break it up into linear factors by using the conjugate rule:


\displaystyle \begin{align} & 3x^{2}-12=3\left( x^{2}-4 \right)=3\left( x+2 \right)\left( x-2 \right) \\ & \\ & x^{2}-1=\left( x+1 \right)\left( x-1 \right) \\ \end{align}

The whole expression is therefore equal to


\displaystyle \frac{3\left( x+2 \right)\left( x-2 \right)\left( x+1 \right)\left( x-1 \right)}{\left( x+1 \right)\left( x+2 \right)}=3\left( x-2 \right)\left( x-1 \right)


NOTE: One can of course expand out the expression to get \displaystyle 3x^{2}-9x+6 as the answer.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.1:5c“