Solution 2.1:4c

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m (Lösning 2.1:4c moved to Solution 2.1:4c: Robot: moved page)
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Instead of multiplying together the whole expression, and then reading off the coefficients, we investigate which terms from the three brackets together give terms in
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<center> [[Image:2_1_4c.gif]] </center>
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<math>x^{1}</math>
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and
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<math>x^{2}</math>.
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If we start with the term in
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<math>x</math>, we see that there is only one combination of a term from each bracket
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which, when multiplied, gives
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<math>x^{1}</math>,
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<math>\left( x-x^{3}+x^{5} \right)\left( 1+3x+5x^{2} \right)\left( 2-7x^{2}-x^{4} \right)=...+x\centerdot 1\centerdot 2+...</math>
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so, the coefficient in front of
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<math>x</math>
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is
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<math>1\centerdot 2=2</math>.
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As for
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<math>x^{2}</math>, we also have only one possible combination:
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<math>\left( x-x^{3}+x^{5} \right)\left( 1+3x+5x^{2} \right)\left( 2-7x^{2}-x^{4} \right)=...+x\centerdot 3x\centerdot 2+...</math>
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The coefficient in front of
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<math>x^{2}</math>
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is
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<math>3\centerdot 2=6</math>

Revision as of 14:36, 15 September 2008

Instead of multiplying together the whole expression, and then reading off the coefficients, we investigate which terms from the three brackets together give terms in \displaystyle x^{1} and \displaystyle x^{2}.

If we start with the term in \displaystyle x, we see that there is only one combination of a term from each bracket which, when multiplied, gives \displaystyle x^{1},


\displaystyle \left( x-x^{3}+x^{5} \right)\left( 1+3x+5x^{2} \right)\left( 2-7x^{2}-x^{4} \right)=...+x\centerdot 1\centerdot 2+...

so, the coefficient in front of \displaystyle x is \displaystyle 1\centerdot 2=2.

As for \displaystyle x^{2}, we also have only one possible combination:


\displaystyle \left( x-x^{3}+x^{5} \right)\left( 1+3x+5x^{2} \right)\left( 2-7x^{2}-x^{4} \right)=...+x\centerdot 3x\centerdot 2+...


The coefficient in front of \displaystyle x^{2} is \displaystyle 3\centerdot 2=6

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