Solution 2.1:2a
From Förberedande kurs i matematik 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, |
| - | < | + | |
| - | {{ | + | {{Displayed math||<math>\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\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | Then, gather together ''x''²-, ''x''- and the constant terms and simplify | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | \phantom{(x-4)(x-5)-3x(2x-3)}&= (x^2-6x^2)+(-5x-4x+9x)+20 \\ | ||
| + | &= -5x^2+0+20\\ | ||
| + | &= \rlap{-5x^2+20\,\textrm{.}}\phantom{x\cdot x-x\cdot 5- 4\cdot x-4\cdot (-5)-(3x \cdot 2x-3x\cdot 3)} | ||
| + | \end{align}</math>}} | ||
Current revision
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 \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\,\textrm{.} \end{align} |
Then, gather together x²-, x- and the constant terms and simplify
| \displaystyle \begin{align}
\phantom{(x-4)(x-5)-3x(2x-3)}&= (x^2-6x^2)+(-5x-4x+9x)+20 \\ &= -5x^2+0+20\\ &= \rlap{-5x^2+20\,\textrm{.}}\phantom{x\cdot x-x\cdot 5- 4\cdot x-4\cdot (-5)-(3x \cdot 2x-3x\cdot 3)} \end{align} |
