Lösung 2.1:4a

Aus Online Mathematik Brückenkurs 1

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K (Lösning 2.1:4a moved to Solution 2.1:4a: Robot: moved page)
Zeile 1: Zeile 1:
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{{NAVCONTENT_START}}
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First, we multiply the second bracket by
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<center> [[Image:2_1_4a.gif]] </center>
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<math>x</math>
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{{NAVCONTENT_STOP}}
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from the first bracket,
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<math>\left( x+2 \right)\left( 3x^{2}-x+5 \right)=x\centerdot 3x^{2}-x\centerdot x+x\centerdot 5+...</math>
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Then, do the same for
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<math>2</math>
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from the first bracket:
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<math>\left( x+2 \right)\left( 3x^{2}-x+5 \right)=3x^{3}-x^{2}+5x+2\centerdot 3x^{2}-2\centerdot x+2\centerdot 5</math>
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Now, collect together
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<math>x^{3}</math>-,
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<math>x^{2}</math>-,
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<math>x</math>- and the constant terms:
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<math>3x^{3}+\left( -1+6 \right)x^{2}+\left( 5-2 \right)x+10=3x^{3}+5x^{2}+3x+10</math>
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The coefficient in front of
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<math>x^{2}</math>
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is
 +
<math>5</math>
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and the coefficient in front of x is
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<math>3</math>.

Version vom 13:42, 15. Sep. 2008

First, we multiply the second bracket by \displaystyle x from the first bracket,


\displaystyle \left( x+2 \right)\left( 3x^{2}-x+5 \right)=x\centerdot 3x^{2}-x\centerdot x+x\centerdot 5+...


Then, do the same for \displaystyle 2 from the first bracket:


\displaystyle \left( x+2 \right)\left( 3x^{2}-x+5 \right)=3x^{3}-x^{2}+5x+2\centerdot 3x^{2}-2\centerdot x+2\centerdot 5


Now, collect together \displaystyle x^{3}-, \displaystyle x^{2}-, \displaystyle x- and the constant terms:


\displaystyle 3x^{3}+\left( -1+6 \right)x^{2}+\left( 5-2 \right)x+10=3x^{3}+5x^{2}+3x+10


The coefficient in front of \displaystyle x^{2} is \displaystyle 5 and the coefficient in front of x is \displaystyle 3.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.1:4a“