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Solution 2.1:4c

From Förberedande kurs i matematik 1

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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 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,

(xx3+x5)(1+3x+5x2)(27x2x4)=+x12+

so, the coefficient in front of x is 12=2.

As for x², we also have only one possible combination

(xx3+x5)(1+3x+5x2)(27x2x4)=+x3x2+

The coefficient in front of x² is 32=6.