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Rational term question

wishmaster

Active member
Oct 11, 2013
211
I have a question why is \(\displaystyle \frac{2x^2-1}{x^2+1}\) equal to \(\displaystyle 2-\frac{3}{x^2+1}\) ??
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
U have a question why is \(\displaystyle \frac{2x^2-1}{x^2+1}\) equal to \(\displaystyle 2-\frac{3}{x^2+1}\) ??
\begin{aligned}
\frac{2x^2-1}{x^2+1}
&= \frac{2x^2 + 2 - 3}{x^2+1} \\
&= \frac{2(x^2 + 1) - 3}{x^2+1} \\
&= \frac{2(x^2 + 1)}{x^2+1} - \frac 3{x^2+1}\\
&= 2-\frac{3}{x^2+1} \\
\end{aligned}
 

wishmaster

Active member
Oct 11, 2013
211
\begin{aligned}
\frac{2x^2-1}{x^2+1}
&= \frac{2x^2 + 2 - 3}{x^2+1} \\
&= \frac{2(x^2 + 1) - 3}{x^2+1} \\
&= \frac{2(x^2 + 1)}{x^2+1} - \frac 3{x^2+1}\\
&= 2-\frac{3}{x^2+1} \\
\end{aligned}
thank you!

Is this the only way to caclulate it?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
thank you!

Is this the only way to caclulate it?
It is similar to calculating:
\begin{aligned}
\frac 7 2
&= \frac {3\cdot 2 + 1}{2} \\
&= \frac {3\cdot 2} 2 + \frac{1}{2} \\
&= 3 + \frac 1 2
\end{aligned}
Just with more complicated expressions.


Alternatively, you can say that if you divide $7$ by $2$, you get $3$ with a remainder of $7 - 3 \cdot 2 = 1$.
That is, \(\displaystyle \frac 7 2 = 3 + \frac 1 2\).


Similarly, you can say that if you divide $2x^2-1$ by $x^2 +1$, you get $2$, because the coefficient of $x^2$ is $2$.

The remainder is then $(2x^2 - 1) - 2(x^2+1) = 2x^2 - 1 - 2x^2 - 2 = -3$.
So
$$\frac{2x^2-1}{x^2+1} = 2 + \frac {-3}{x^2+1}$$

Note that this "other" method is really the same thing.
 

soroban

Well-known member
Feb 2, 2012
409
Hello, wishmaster!

[tex]\text{Why is }\,\frac{2x^2-1}{x^2+1}\,\text{ equal to }\,2-\frac{3}{x^2+1}\,?[/tex]

Are you familiar with Long Division?

. . [tex]\begin{array}{cccccc}
&&&& 2 \\
&& --&--&-- \\
x^2+1 & | & 2x^2 &-&1 \\
&& 2x^2 &+& 2 \\
&& --&--&-- \\
&&& - & 3 \end{array}[/tex]


Therefore: .[tex]\frac{2x^2-1}{x^2+1} \;\;=\;\;2 - \frac{3}{x^2+1}[/tex]
 

wishmaster

Active member
Oct 11, 2013
211
Hello, wishmaster!


Are you familiar with Long Division?

. . [tex]\begin{array}{cccccc}
&&&& 2 \\
&& --&--&-- \\
x^2+1 & | & 2x^2 &-&1 \\
&& 2x^2 &+& 2 \\
&& --&--&-- \\
&&& - & 3 \end{array}[/tex]


Therefore: .[tex]\frac{2x^2-1}{x^2+1} \;\;=\;\;2 - \frac{3}{x^2+1}[/tex]

Thank you all for the help! I have studied those kind of problems,so now i know how to do it!