Sorry if I call something by the wrong name since I didnt learn math in english. ok so for example this: (a+b)(a-b) if you break it down to the second "()" you will end up with this: a+-b could somebody explain why + shows up?
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1$\begingroup$ What do you mean by "if you break it down to the second ()"? $\endgroup$– HakimOct 22, 2014 at 17:48
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$\begingroup$ Do you mean getting rid of the parentheses? $\endgroup$– AdnanOct 22, 2014 at 17:48
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$\begingroup$ yeah, yeah sorry $\endgroup$– NikkoOct 22, 2014 at 17:49
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1 Answer
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The method to use is called factorisation:
To clear things up, let's give the variables a color: $(\color{blue}{a}\color{red}{+b})(\color{blue}{a}\color{red}{-b})$. We need to apply the following rule:
$\color{blue}{a}$ is multiplied first with $\color{blue}{a}$ and then with $\color{red}{-b}$ and we get the following: $\color{blue}{aa}$ and $\color{red}{-}\color{blue}{a}\color{red}{b}$. Then we multiply $\color{red}{+b}$ with $\color{blue}{a}$ and then $\color{red}{-b}$, we get the following: $\color{red}{b}\color{blue}{a}$ and $\color{red}{-bb}$. Adding these up will get you:
$(\color{blue}{aa})+(\color{red}{-}\color{blue}{a}\color{red}{b})+(\color{red}{b}\color{blue}{a})+(\color{red}{-bb})$. Since $x+-y = x-y$, we get $\color{blue}{aa}\color{red}{-}\color{blue}{a}\color{red}{b}+\color{red}{b}\color{blue}{a}\color{red}{-bb}$.
Also $x\cdot x=x^2$ and $p\cdot q=q\cdot p$, we get $a^2-ab+ab-b^2 = \color{blue}{a^2}-\color{red}{b^2}$
