Lösung 2.1:2b

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We expand the first product of bracketed terms by multiplying each term inside the first bracket by each term from the second bracket
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<math>
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\qquad
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\begin{align}
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(1-5x)(1+15x) &= 1\cdot 1+1\cdot 15x-5x\cdot 1-5x \cdot 15x\\
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&=1+15x-5x-75x^2
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\end{align}
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</math>
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As for the second expression, we can use the conjugate rule <math>(a-b)(a+b)=a^2-b^2,</math> where <math>a=2</math> and <math> b=5x.</math>
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<math>
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\qquad
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\begin{align}
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3(2-5x)(2+5x) &= 3\big( 2^2-(5x)^2\big)\\
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&=3(4-25x^2)\\
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&=12-75x^2
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\end{align}
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</math>
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All together, we obtain
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<math> \qquad (1-5x)(1+15x)-3(2-5x)(2+5x) </math>
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<math>
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\qquad
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\begin{align}
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\phantom{3(2-5x)(2+5x)} &= (1+10x-75x^2)-(12-75x^2)\\
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&= 1+10x-75x^2-12+75x^2\\
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&= 1-12+10x-75x^2+75x^2\\
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&=-11+10x
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\end{align}
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</math>
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Version vom 10:58, 13. 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}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.1:2b“