Solution 2.1:2b
From Förberedande kurs i matematik 1
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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 |
| - | < | + | |
| - | {{ | + | {{Displayed math||<math>\begin{align} |
| + | (1-5x)(1+15x) &= 1\cdot 1+1\cdot 15x-5x\cdot 1-5x \cdot 15x\\ | ||
| + | &=1+15x-5x-75x^2\\ | ||
| + | &=1+10x-75x^2\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | 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>, | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | 3(2-5x)(2+5x) &= 3\big( 2^2-(5x)^2\big)\\ | ||
| + | &=3(4-25x^2)\\ | ||
| + | &=12-75x^2\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | All together, we obtain | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | (1-5x)(1+15x)-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\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
Current revision
We expand the first product of bracketed terms by multiplying each term inside the first bracket by each term from the second bracket
| \displaystyle \begin{align}
(1-5x)(1+15x) &= 1\cdot 1+1\cdot 15x-5x\cdot 1-5x \cdot 15x\\ &=1+15x-5x-75x^2\\ &=1+10x-75x^2\,\textrm{.} \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 \begin{align}
3(2-5x)(2+5x) &= 3\big( 2^2-(5x)^2\big)\\ &=3(4-25x^2)\\ &=12-75x^2\,\textrm{.} \end{align} |
All together, we obtain
| \displaystyle \begin{align}
(1-5x)(1+15x)-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\,\textrm{.} \end{align} |
