Lösung 2.1:2d
Aus Online Mathematik Brückenkurs 1
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Version vom 13:44, 22. Okt. 2008
The three brackets can be multiplied together in any order, but just in this case it seems appropriate to start by multiplying the first two brackets because we can carry out the multiplication directly by using the conjugate rule,
| \displaystyle \begin{align}
(3x^2+2)(3x^2-2)(9x^4+4) &= \big( (3x^2)^2-2^2 \big) (9x^4+4)\\ &= (9x^4-4)(9x^4+4)\,\textrm{.} \end{align} |
We happend to obtain an expression which can also be multiplied using the conjugate rule,
| \displaystyle (9x^4-4)(9x^4+4)=(9x^4)^2-4^2=81x^8-16\,\textrm{.} |
Note: If we had instead multiplied together the terms in the first and third brackets first, we would also have got the right answer, but the calculations would have been lengthier.
