Lösung 2.1:4c

Aus Online Mathematik Brückenkurs 1

Version vom 13:47, 22. Okt. 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 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,

(x−x3+x5)(1+3x+5x2)(2−7x2−x4)=+x12+

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

As for x², we also have only one possible combination

(x−x3+x5)(1+3x+5x2)(2−7x2−x4)=+x3x2+

The coefficient in front of x² is 32=6.