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zaphodBest ResponseYou've already chosen the best response.0
\[(x2)(\frac{100}{x}  5)\]
 one year ago

zaphodBest ResponseYou've already chosen the best response.0
@ParthKohli @AccessDenied @Aadarsh
 one year ago

nbouscalBest ResponseYou've already chosen the best response.2
\[\begin{align} \text{F.O.I.L. = First Outside Inside Last}\\ \hline \\ &(ax+b)(cx+d)\\ \ \\ \text{First:}&(\color{blue}{ax}+b)(\color{blue}{cx}+d)\to \color{blue}{ax}\cdot\color{blue}{cx}=\color{blue}{acx^2}\\ \text{Outside:}&(\color{green}{ax}+b)(cx+\color{green}{d})\to \color{green}{ax}\cdot\color{green}{d}=\color{green}{adx}\\ \text{Inside:}&(ax+\color{red}{b})(\color{red}{cx}+d)\to \color{red}{b}\cdot\color{red}{cx}=\color{red}{bcx}\\ \text{Last:}&(ax+\color{orange}{b})(cx+\color{orange}{d})\to \color{orange}{b}\cdot\color{orange}{d}=\color{orange}{bd}\\ \hline \\ (ax+b)(cx+d)&=\color{blue}{acx^2}+\color{green}{adx}+\color{red}{bcx}+\color{orange}{bd} \end{align}\] Only difference from a typical FOIL is that you have a 1/x term, but you can still use the same exact method.
 one year ago

AadarshBest ResponseYou've already chosen the best response.0
@zaphod , please proceed as instructed by @nbouscal , u will get the answer. Its easy.
 one year ago

zaphodBest ResponseYou've already chosen the best response.0
wow, i like this method :D, thanks alot @nbouscal
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.2
\(\begin{align} \text{F.O.I.L. = First Outside Inside Last}\\ \hline \\ &(ax+b)(cx+d)\\ \ \\ \text{First:}&(\color{blue}{ax}+b)(\color{blue}{cx}+d)\to \color{blue}{ax}\cdot\color{blue}{cx}=\color{blue}{acx^2}\\ \text{Outside:}&(\color{green}{ax}+b)(cx+\color{green}{d})\to \color{green}{ax}\cdot\color{green}{d}=\color{green}{adx}\\ \text{Inside:}&(ax+\color{red}{b})(\color{red}{cx}+d)\to \color{red}{b}\cdot\color{red}{cx}=\color{red}{bcx}\\ \text{Last:}&(ax+\color{orange}{b})(cx+\color{orange}{d})\to \color{orange}{b}\cdot\color{orange}{d}=\color{orange}{bd}\\ \hline \\ (ax+b)(cx+d)&=\color{blue}{acx^2}+\color{green}{adx}+\color{red}{bcx}+\color{orange}{bd} \end{align}\)
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.2
Lol nathan you copied me
 one year ago

zaphodBest ResponseYou've already chosen the best response.0
what about this method (ax+b)(cx+d) cx(ax+b)+d(ax+b)> very simple :)
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.2
Of course...nathan uses foil..I use the method you stated
 one year ago

zaphodBest ResponseYou've already chosen the best response.0
oh any other methods? easier than this
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.2
That's it....all of the methods have the same intuition
 one year ago

nbouscalBest ResponseYou've already chosen the best response.2
They're the same method really, just different ways of looking at it. I actually prefer to use the long multiplication method, because it works for multiplying trinomials and polynomials as well.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.2
Yeah, that's what I was meaning
 one year ago

nbouscalBest ResponseYou've already chosen the best response.2
Example:\[ \begin{align} &&&x+2\\ &\times&&\color{red}{x}+\color{blue}{1}\\ \hline \\ &&&\color{blue}{x+2}\\ &+&\color{red}{x^2+2}&\color{red}{x}\\ \hline \\ &&x^2+3&x+2 \end{align} \]Just like long multiplying numbers, but with polynomials instead. That's my preferred method :)
 one year ago
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