Lösung 2.1:2b
Aus Online Mathematik Brückenkurs 1
K (Robot: Automated text replacement (-[[Bild: +[[Image:)) |
|||
| Zeile 1: | Zeile 1: | ||
{{NAVCONTENT_START}} | {{NAVCONTENT_START}} | ||
| - | <!--center> [[ | + | <!--center> [[Image:2_1_2b.gif]] </center--> |
We expand the first product of bracketed terms by multiplying each term inside the first bracket by each term from the second bracket | We expand the first product of bracketed terms by multiplying each term inside the first bracket by each term from the second bracket | ||
Version vom 06:33, 21. Aug. 2008
We expand the first product of bracketed terms by multiplying each term inside the first bracket by each term from the second bracket
\displaystyle \qquad \begin{align} (1-5x)(1+15x) &= 1\cdot 1+1\cdot 15x-5x\cdot 1-5x \cdot 15x\\ &=1+15x-5x-75x^2 \end{align}
As for the second expression, we can use the conjugate rule \displaystyle (a-b)(a+b)=a^2-b^2, where \displaystyle a=2 and \displaystyle b=5x.
\displaystyle \qquad \begin{align} 3(2-5x)(2+5x) &= 3\big( 2^2-(5x)^2\big)\\ &=3(4-25x^2)\\ &=12-75x^2 \end{align}
All together, we obtain
\displaystyle \qquad (1-5x)(1+15x)-3(2-5x)(2+5x)
\displaystyle \qquad \begin{align} \phantom{3(2-5x)(2+5x)} &= (1+10x-75x^2)-(12-75x^2)\\ &= 1+10x-75x^2-12+75x^2\\ &= 1-12+10x-75x^2+75x^2\\ &=-11+10x \end{align}
