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zaphod

  • 2 years ago

how do i simplify thisl..

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  1. zaphod
    • 2 years ago
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    \[(x-2)(\frac{100}{x} - 5)\]

  2. zaphod
    • 2 years ago
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    @ParthKohli @AccessDenied @Aadarsh

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

  4. Aadarsh
    • 2 years ago
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    @zaphod , please proceed as instructed by @nbouscal , u will get the answer. Its easy.

  5. zaphod
    • 2 years ago
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    wow, i like this method :D, thanks alot @nbouscal

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

  7. ParthKohli
    • 2 years ago
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    Lol nathan you copied me

  8. zaphod
    • 2 years ago
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    what about this method (ax+b)(cx+d) cx(ax+b)+d(ax+b)--------> very simple :)

  9. ParthKohli
    • 2 years ago
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    Of course...nathan uses foil..I use the method you stated

  10. zaphod
    • 2 years ago
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    oh any other methods? easier than this

  11. ParthKohli
    • 2 years ago
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    That's it....all of the methods have the same intuition

  12. nbouscal
    • 2 years ago
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    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
    • 2 years ago
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    Yeah, that's what I was meaning

  14. nbouscal
    • 2 years ago
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    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 :)

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