Lösung 2.1:2a

Aus Online Mathematik Brückenkurs 1

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First, multiply the brackets together. In the first product, every term in the first bracket is multiplied by every term in the second bracket,
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<math>
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\qquad
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\begin{align}
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(x-4)(x-5)-3x(2x-3)&= x\cdot x-x\cdot 5- 4\cdot x-4\cdot (-5)-(3x \cdot 2x-3x\cdot 3)\\
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&= x^2-5x-4x+20-(6x^2-9x)\\
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&=x^2-5x-4x+20-6x^2+9x
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\end{align}
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</math>
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Then, gather together <math>x^2-, \, x- </math> and the constant terms and simplify
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<math>
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\qquad
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\begin{align}
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\phantom{(x-4)(x-5)-3x(2x-3)}&= (x^2-6x^2)+(-5x-4x+9x)+20 \\
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&= -5x^2+0+20\\
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&= -5x^2+20
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\end{align}
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</math>
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Version vom 10:01, 13. Aug. 2008

Bild:2_1_2a.gif

First, multiply the brackets together. In the first product, every term in the first bracket is multiplied by every term in the second bracket,

\displaystyle \qquad \begin{align} (x-4)(x-5)-3x(2x-3)&= x\cdot x-x\cdot 5- 4\cdot x-4\cdot (-5)-(3x \cdot 2x-3x\cdot 3)\\ &= x^2-5x-4x+20-(6x^2-9x)\\ &=x^2-5x-4x+20-6x^2+9x \end{align}

Then, gather together \displaystyle x^2-, \, x- and the constant terms and simplify

\displaystyle \qquad \begin{align} \phantom{(x-4)(x-5)-3x(2x-3)}&= (x^2-6x^2)+(-5x-4x+9x)+20 \\ &= -5x^2+0+20\\ &= -5x^2+20 \end{align}

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