From Förberedande kurs i matematik 1

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-	{{NAVCONTENT_START}}	+	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.
-	<center> [[Image:2_1_4c.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	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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		+	{{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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		+	so, the coefficient in front of ''x'' is <math>1\cdot 2 = 2\,</math>.
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		+	As for ''x''², we also have only one possible combination
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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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		+	The coefficient in front of ''x''² is <math>3\cdot 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\,.