Question

how do i simplify thisl..

Answer 1

\[(x-2)(\frac{100}{x} - 5)\]

Answer 2

@ParthKohli @AccessDenied @Aadarsh

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

\(\begin{align}
\text{F.O.I.L. = First Outside Inside Last}\\
\hline \\
&(ax+b)(cx+d)\\ \ \\
\text{First:}&(\color{blue}{ax}+b)(\color{blue}{cx}+d)\to \color{blue}{ax}\cdot\color{blue}{cx}=\color{blue}{acx^2}\\
\text{Outside:}&(\color{green}{ax}+b)(cx+\color{green}{d})\to \color{green}{ax}\cdot\color{green}{d}=\color{green}{adx}\\
\text{Inside:}&(ax+\color{red}{b})(\color{red}{cx}+d)\to \color{red}{b}\cdot\color{red}{cx}=\color{red}{bcx}\\
\text{Last:}&(ax+\color{orange}{b})(cx+\color{orange}{d})\to \color{orange}{b}\cdot\color{orange}{d}=\color{orange}{bd}\\
\hline \\
(ax+b)(cx+d)&=\color{blue}{acx^2}+\color{green}{adx}+\color{red}{bcx}+\color{orange}{bd}
\end{align}\)

Answer 7

Lol nathan you copied me

Answer 8

what about this method
(ax+b)(cx+d)
cx(ax+b)+d(ax+b)--------> very simple :)

Answer 9

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

Answer 10

oh any other methods? easier than this

Answer 11

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

Answer 12

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.

Answer 13

Yeah, that's what I was meaning

Answer 14

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 :)