rational expressions #4
- Thread starter mathymath
- Start date
Hello, mathymath!
Those "trinomials" do not factor.
\(\displaystyle \text{We have: }\:\dfrac{x^2+x+2}{x+1} - \dfrac{x}{1} \:=\:\dfrac{x^2-5}{(x-1)(x+1)}\)
\(\displaystyle \text{Multiply by }(x-1)(x+1)\!:\)
. . \(\displaystyle \begin{array}{cc}(x-1)(x^2+x+2) - x(x-1)(x+1) \;=\; x^2-5 \\ \\ x^3 + x^2 + 2x - x^2 - x - 2 -x^3 + x \;=\; x^2-5 \\ \\ x^2 - 2x - 3 \;=\;0 \\ \\ (x-3)(x+1) \;=\;0 \\ \\ x\:=\:3,\,\color{red}{\rlap{//}}\text{-}1 \end{array}\)
\(\displaystyle \text{Therefore: }\:x \:=\:3\)
Those "trinomials" do not factor.
Note that: \(\displaystyle x \,\ne\,\pm1\)
\(\displaystyle \text{We have: }\:\dfrac{x^2+x+2}{x+1} - \dfrac{x}{1} \:=\:\dfrac{x^2-5}{(x-1)(x+1)}\)
\(\displaystyle \text{Multiply by }(x-1)(x+1)\!:\)
. . \(\displaystyle \begin{array}{cc}(x-1)(x^2+x+2) - x(x-1)(x+1) \;=\; x^2-5 \\ \\ x^3 + x^2 + 2x - x^2 - x - 2 -x^3 + x \;=\; x^2-5 \\ \\ x^2 - 2x - 3 \;=\;0 \\ \\ (x-3)(x+1) \;=\;0 \\ \\ x\:=\:3,\,\color{red}{\rlap{//}}\text{-}1 \end{array}\)
\(\displaystyle \text{Therefore: }\:x \:=\:3\)
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You've said elsewhere that copying solutions is how you learn. Does this mean that you now understand this topic?
HallsofIvy
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I am sorry to hear you say that. You learn math by doing math, not by seeing someone else do it! It is very easy, all to common, for a person to watch another solve a math problem and, because the step are so logical, think "Oh, yes, I understand that now". But when they try to do another problem by themselves, cannot do it. Yes, looking at one or two examples will help you see what to do but until you have done several yourself, right from the start, you do NOT know how to do the problems. If you want to compare your work to someone else's, start by showing what you did first.