Lösung 2.1:5d

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The factor <math>y^{2}+4y+4</math> can be rewritten as <math>y^{2}+2\cdot 2y+2^{2}</math>, which opens the way for using the squaring rule
The factor <math>y^{2}+4y+4</math> can be rewritten as <math>y^{2}+2\cdot 2y+2^{2}</math>, which opens the way for using the squaring rule
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{{Displayed math||<math>y^{2}+4y+4 = y^{2}+2\cdot 2y+2^{2} = (y+2)^{2}\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>y^{2}+4y+4 = y^{2}+2\cdot 2y+2^{2} = (y+2)^{2}\textrm{.}</math>}}
The factor <math>2y-4</math> is already a first-order expression and can therefore not be divided up any further, other than by taking out a factor <math>2</math>, i.e. <math>2y-4=2\left( y-2 \right)\,</math>.
The factor <math>2y-4</math> is already a first-order expression and can therefore not be divided up any further, other than by taking out a factor <math>2</math>, i.e. <math>2y-4=2\left( y-2 \right)\,</math>.
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<math>y^{2}-4</math> can be factorized using the conjugate rule to give
<math>y^{2}-4</math> can be factorized using the conjugate rule to give
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{{Displayed math||<math>y^{2}-4 = (y+2)(y-2)\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>y^{2}-4 = (y+2)(y-2)\,\textrm{.}</math>}}
On the other hand, <math>y^{2}+4</math> cannot be written as a product of first-order factors. If it were possible to write <math>y^{2}+4 = (y-a)(y-b)</math>, where ''a'' and ''b'' are some numbers, then <math>y=a</math> and
On the other hand, <math>y^{2}+4</math> cannot be written as a product of first-order factors. If it were possible to write <math>y^{2}+4 = (y-a)(y-b)</math>, where ''a'' and ''b'' are some numbers, then <math>y=a</math> and
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Thus,
Thus,
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{{Displayed math||<math>\frac{(y^{2}+4y+4)(2y-4)}{(y^{2}+4)(y^{2}-4)} = \frac{(y+2)^{2}\cdot 2(y-2)}{(y^{2}+4)(y+2)(y-2)} = \frac{2(y+2)}{(y^{2}+4)}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\frac{(y^{2}+4y+4)(2y-4)}{(y^{2}+4)(y^{2}-4)} = \frac{(y+2)^{2}\cdot 2(y-2)}{(y^{2}+4)(y+2)(y-2)} = \frac{2(y+2)}{(y^{2}+4)}\,\textrm{.}</math>}}

Version vom 08:24, 22. Okt. 2008

In this fraction, there is the possibility that the numerator and denominator contain common factors which can be eliminated and we therefore try to factorize all expressions to simplest possible form.

The factor \displaystyle y^{2}+4y+4 can be rewritten as \displaystyle y^{2}+2\cdot 2y+2^{2}, which opens the way for using the squaring rule

\displaystyle y^{2}+4y+4 = y^{2}+2\cdot 2y+2^{2} = (y+2)^{2}\textrm{.}

The factor \displaystyle 2y-4 is already a first-order expression and can therefore not be divided up any further, other than by taking out a factor \displaystyle 2, i.e. \displaystyle 2y-4=2\left( y-2 \right)\,.


\displaystyle y^{2}-4 can be factorized using the conjugate rule to give

\displaystyle y^{2}-4 = (y+2)(y-2)\,\textrm{.}

On the other hand, \displaystyle y^{2}+4 cannot be written as a product of first-order factors. If it were possible to write \displaystyle y^{2}+4 = (y-a)(y-b), where a and b are some numbers, then \displaystyle y=a and \displaystyle y=b would be zeros of \displaystyle y^{2}+4, but because \displaystyle y^{2}+4 is the sum of a square, \displaystyle y^{2}, which cannot have a negative value and the number \displaystyle 4, \displaystyle y^{2}+4 is always greater than or equal \displaystyle 4 regardless of how \displaystyle y is chosen. Hence, \displaystyle y^{2}+4 cannot be divided up into first-order factors.

Thus,

\displaystyle \frac{(y^{2}+4y+4)(2y-4)}{(y^{2}+4)(y^{2}-4)} = \frac{(y+2)^{2}\cdot 2(y-2)}{(y^{2}+4)(y+2)(y-2)} = \frac{2(y+2)}{(y^{2}+4)}\,\textrm{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.1:5d“