Algebraic expression #6

[mathwannabe]

Hello everyone

I just managed to finish one problem I think was very hard (after bashing the wall with my head for a god few times). I mean, come on... Is this supposed to be a 1st year of high school level???

1) \(\displaystyle \dfrac{x^4+x^2+1}{x^4+3x^2+2x^3+2x+1}=\)

\(\displaystyle =\dfrac{x^4+x^2+1}{x^4+2x^3+3x^2+2x+1}=\)

\(\displaystyle =\dfrac{x^4+x^2+1}{x^4+x^2+1+2x^3+2x^2+2x}=\)

\(\displaystyle =\dfrac{x^4+x^2+1}{(x^2+x+1)^2}=\)

Like fixing that denominator wasn't hard enough, the real torture started now x)

\(\displaystyle =\dfrac{x^4+2x^2+1-x^2}{(x^2+x+1)^2}=\)

\(\displaystyle =\dfrac{(x^2+1)^2-x^2}{(x^2+x+1)^2}=\)

\(\displaystyle =\dfrac{(x^2+1-x)(x^2+1+x)}{(x^2+x+1)^2}=\)

\(\displaystyle =\dfrac{x^2+1-x}{x^2+x+1}\)

 

[Deleted member 4993]

hello everyone :d

i just managed to finish one problem i think was very hard (after bashing the wall with my head for a god few times). I mean, come on... Is this supposed to be a 1st year of high school level???

1) \(\displaystyle \dfrac{x^4+x^2+1}{x^4+3x^2+2x^3+2x+1}=\)

\(\displaystyle =\dfrac{x^4+x^2+1}{x^4+2x^3+3x^2+2x+1}=\)

\(\displaystyle =\dfrac{x^4+x^2+1}{x^4+x^2+1+2x^3+2x^2+2x}=\)

\(\displaystyle =\dfrac{x^4+x^2+1}{(x^2+x+1)^2}=\)

like fixing that denominator wasn't hard enough, the real torture started now x)

\(\displaystyle =\dfrac{x^4+2x^2+1-x^2}{(x^2+x+1)^2}=\)

\(\displaystyle =\dfrac{(x^2+1)^2-x^2}{(x^2+x+1)^2}=\)

\(\displaystyle =\dfrac{(x^2+1-x)(x^2+1+x)}{x^2+x+1}=\)

\(\displaystyle =\dfrac{x^2+1-x}{x^2+x+1}\) <<<< this step is incorrect - look at it carefully....

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[mathwannabe]

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Professor SK, OP edited to fix the mistake. Was a typo.

 

[mathwannabe]

I agree that few high school students are going to hack their way through this. It is, I suppose, good that the teacher can do problems that are too hard for most of the students. The opposite might be awkward.

Ha ha... Yes, I think that the author (who is a living legend and who is infamous for making the lives of high school students miserable) was heavily intoxicated when he put THAT into a high school collection of problems book. Just joking ofc, I know that they like putting problems like THAT in order to find those rare precious few who are gifted for this stuff. Which I am NOT, but I guess stubbornness sometimes pays off

 

[soroban]

Hello, mathwannabe!

This is definitely NOT a first-year problem.
I factored and reduced it, thanks to years of experience ... and a devious imagination.

\(\displaystyle (1)\;\dfrac{x^4+x^2+1}{x^4+2x^3 + 3x^2 + 2x + 1}\)

The numerator is: .\(\displaystyle x^4 + 2x^2 + 1 - x^2 \:=\x^2+1)^2 - x^2 \)

. . . . . . . . . . . \(\displaystyle =\;(x^2+1 - x)(x^2+1+x) \:=\x^2-x+1)(x^2+x+1)\)


The denominator is: .\(\displaystyle (x^4 + x^3 + x^2) + (x^3 + x^2 + x) + (x^2 + x + 1) \)

. . . . . . . . . . . . .\(\displaystyle =\;x^2(x^2+x + 1) + x(x^2 + x + 1) + (x^2 + x + 1)\)

. . . . . . . . . . . . .\(\displaystyle =\;(x^2 + x + 1)(x^2 + x + 1)\)


The fraction becomes: .\(\displaystyle \dfrac{(x^2-x+1)(x^2+x+1)}{(x^2+x+1)(x^2+x+1)} \;=\;\dfrac{x^2-x+1}{x^2+x+1} \)

 

[mathwannabe]

Hello, mathwannabe!

This is definitely NOT a first-year problem.
I factored and reduced it, thanks to years of experience ... and a devious imagination.


The numerator is: .\(\displaystyle x^4 + 2x^2 + 1 - x^2 \:=\x^2+1)^2 - x^2 \)

. . . . . . . . . . . \(\displaystyle =\;(x^2+1 - x)(x^2+1+x) \:=\x^2-x+1)(x^2+x+1)\)


The denominator is: .\(\displaystyle (x^4 + x^3 + x^2) + (x^3 + x^2 + x) + (x^2 + x + 1) \)

. . . . . . . . . . . . .\(\displaystyle =\;x^2(x^2+x + 1) + x(x^2 + x + 1) + (x^2 + x + 1)\)

. . . . . . . . . . . . .\(\displaystyle =\;(x^2 + x + 1)(x^2 + x + 1)\)


The fraction becomes: .\(\displaystyle \dfrac{(x^2-x+1)(x^2+x+1)}{(x^2+x+1)(x^2+x+1)} \;=\;\dfrac{x^2-x+1}{x^2+x+1} \)

Hello soroban. Thank you for reply.

 

[mathwannabe]

Hello, mathwannabe!

This is definitely NOT a first-year problem.
I factored and reduced it, thanks to years of experience ... and a devious imagination.

The denominator is: .\(\displaystyle (x^4 + x^3 + x^2) + (x^3 + x^2 + x) + (x^2 + x + 1) \)

. . . . . . . . . . . . .\(\displaystyle =\;x^2(x^2+x + 1) + x(x^2 + x + 1) + (x^2 + x + 1)\)

. . . . . . . . . . . . .\(\displaystyle =\;(x^2 + x + 1)(x^2 + x + 1)\)

I actually like how you dealt with the denominator there a lot more then how I did it.