Aus Online Mathematik Brückenkurs 1

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