## zaphod 3 years ago how do i simplify thisl..

1. zaphod

$(x-2)(\frac{100}{x} - 5)$

2. zaphod

3. nbouscal

\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.

@zaphod , please proceed as instructed by @nbouscal , u will get the answer. Its easy.

5. zaphod

wow, i like this method :D, thanks alot @nbouscal

6. ParthKohli

\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}

7. ParthKohli

Lol nathan you copied me

8. zaphod

9. ParthKohli

Of course...nathan uses foil..I use the method you stated

10. zaphod

oh any other methods? easier than this

11. ParthKohli

That's it....all of the methods have the same intuition

12. nbouscal

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.

13. ParthKohli

Yeah, that's what I was meaning

14. nbouscal

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 :)