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Lösung 2.1:5d

Aus Online Mathematik Brückenkurs 1

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K (Lösning 2.1:5d moved to Solution 2.1:5d: Robot: moved page)
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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.
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<center> [[Image:2_1_5d-1(2).gif]] </center>
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The factor
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<math>y^{2}+4y+4</math>
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<center> [[Image:2_1_5d-2(2).gif]] </center>
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can be rewritten as
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<math>y^{2}+2\centerdot 2y+2^{2}</math>, which opens the way for using the squaring rule
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<math>y^{2}+4y+4=y^{2}+2\centerdot 2y+2^{2}=\left( y+2 \right)^{2}</math>
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The factor
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<math>2y-4</math>
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is already a first-order expression and can therefore not be divided up any further,
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other than by taking out a factor
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<math>2</math>
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i.e.
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<math>2y-4=2\left( y-2 \right)</math>
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.
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<math>y^{2}+4</math>
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y2-4 can be factorized using the conjugate rule to give
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<math>y^{2}-4=\left( y+2 \right)\left( y-2 \right)</math>
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On the other hand,
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<math>y^{2}+4</math>
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cannot be written as a product of first-order factors. If it were possible to
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write
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<math>y^{2}+4=\left( y-a \right)\left( y-b \right)</math>, where
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<math>a</math>
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and
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<math>b</math>
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are some numbers, then
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<math>y=a</math>
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and
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<math>y=b</math>
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would be zeros of
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<math>y^{2}+4</math>, but because
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<math>y^{2}+4</math>
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is the sum of a square,
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<math>y^{2}</math>, which cannot have a negative value and the number
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<math>4</math>,
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<math>y^{2}+4</math>
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is always greater than or equal
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<math>4</math>
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regardless of how
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<math>y</math>
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is chosen. Hence,
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<math>y^{2}+4</math>
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cannot be divided up into first-order factors.
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Thus,
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<math>\frac{\left( y^{2}+4y+4 \right)\left( 2y-4 \right)}{\left( y^{2}+4 \right)\left( y^{2}-4 \right)}=\frac{\left( y+2 \right)^{2}\centerdot 2\left( y-2 \right)}{\left( y^{2}+4 \right)\left( y+2 \right)\left( y-2 \right)}=\frac{2\left( y+2 \right)}{\left( y^{2}+4 \right)}</math>

Version vom 09:55, 16. Sep. 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 y2+4y+4 can be rewritten as y2+22y+22, which opens the way for using the squaring rule


y2+4y+4=y2+22y+22=y+22 


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


y2+4 y2-4 can be factorized using the conjugate rule to give


y2−4=y+2y−2 


On the other hand, y2+4 cannot be written as a product of first-order factors. If it were possible to write y2+4=y−ay−b , where a and b are some numbers, then y=a and y=b would be zeros of y2+4, but because y2+4 is the sum of a square, y2, which cannot have a negative value and the number 4, y2+4 is always greater than or equal 4 regardless of how y is chosen. Hence, y2+4 cannot be divided up into first-order factors.

Thus,


y2+4y2−4y2+4y+42y−4=y+222y−2y2+4y+2y−2=y2+42y+2 

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