Why is (2x^2)/(x^2+1) not dividing evenly?

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The discussion centers on the polynomial division of (2x^2)/(x^2+1). The user initially misunderstands the division process, mistakenly placing the divisor incorrectly. The correct approach involves performing polynomial long division with the divisor (x^2 + 1) outside the division symbol, leading to a quotient of 2 + (-2)/(x^2 + 1). This method confirms that multiplying the quotient by the divisor yields the original numerator, 2x^2.

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I've looked examples up online and I just can't figure out what to do exactly when I have (2x^2)/(x^2+1), for some stupid reason that was probably the work of satan, EVERY problem on the internet only has the lead coefficient of the numerator equal to or less than that in the denominator and when I divide I just don't get the answer I am suppose to get.

...(1/2)?
2x^2...|(x^2+1)``````
2x^2 goes into x^2 one half times according to my mathematics, but somehow that answer doesn't work, somehow 2x^2 goes into x^2 twice.
 
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questionasker1 said:
I've looked examples up online and I just can't figure out what to do exactly when I have (2x^2)/(x^2+1), for some stupid reason that was probably the work of satan, EVERY problem on the internet only has the lead coefficient of the numerator equal to or less than that in the denominator and when I divide I just don't get the answer I am suppose to get.

...(1/2)?
2x^2...|(x^2+1)``````
2x^2 goes into x^2 one half times according to my mathematics, but somehow that answer doesn't work, somehow 2x^2 goes into x^2 twice.
You're doing the division backwards. The divisor is x2 + 1, so it should be outside when you do the long division. Your quotient should be 2 + (-2)/(x2 + 1).

As a check, multiply the quotient ( 2 + (-2)/(x2 + 1) ) by the divisor ( x2 + 1 ), and you'll get 2x2.
 
Ok that seemed to solve it, thanks.
 

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