Solution 2.1:4c

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Current revision (10:26, 23 September 2008) (edit) (undo)
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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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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 ''x''¹ and ''x''².
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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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If we start with the term in ''x'', we see that there is only one combination of a term from each bracket which, when multiplied, gives ''x''¹,
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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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{{Displayed math||<math>(\underline{x}-x^{3}+x^{5})(\underline{1}+3x+5x^{2})(\underline{2}-7x^{2}-x^{4}) = \cdots + \underline{x\cdot 1\cdot 2} + \cdots</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 ''x'' is <math>1\cdot 2 = 2\,</math>.
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so, the coefficient in front of
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As for ''x''², we also have only one possible combination
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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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{{Displayed math||<math>(\underline{x}-x^{3}+x^{5})(1+\underline{3x}+5x^{2})(\underline{2}-7x^{2}-x^{4}) = \cdots + \underline{x\cdot 3x\cdot 2} + \cdots</math>}}
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<math>x^{2}</math>, we also have only one possible combination:
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The coefficient in front of ''x''² is <math>3\cdot 2 = 6\,</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 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>
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Current revision

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 x¹ and x².

If we start with the term in x, we see that there is only one combination of a term from each bracket which, when multiplied, gives x¹,

\displaystyle (\underline{x}-x^{3}+x^{5})(\underline{1}+3x+5x^{2})(\underline{2}-7x^{2}-x^{4}) = \cdots + \underline{x\cdot 1\cdot 2} + \cdots

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

As for x², we also have only one possible combination

\displaystyle (\underline{x}-x^{3}+x^{5})(1+\underline{3x}+5x^{2})(\underline{2}-7x^{2}-x^{4}) = \cdots + \underline{x\cdot 3x\cdot 2} + \cdots

The coefficient in front of x² is \displaystyle 3\cdot 2 = 6\,.

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