# Rational term question

#### wishmaster

##### Active member
I have a question why is $$\displaystyle \frac{2x^2-1}{x^2+1}$$ equal to $$\displaystyle 2-\frac{3}{x^2+1}$$ ??

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#### Klaas van Aarsen

##### MHB Seeker
Staff member
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
\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
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
Hello, wishmaster!

$$\text{Why is }\,\frac{2x^2-1}{x^2+1}\,\text{ equal to }\,2-\frac{3}{x^2+1}\,?$$

Are you familiar with Long Division?

. . $$\begin{array}{cccccc} &&&& 2 \\ && --&--&-- \\ x^2+1 & | & 2x^2 &-&1 \\ && 2x^2 &+& 2 \\ && --&--&-- \\ &&& - & 3 \end{array}$$

Therefore: .$$\frac{2x^2-1}{x^2+1} \;\;=\;\;2 - \frac{3}{x^2+1}$$

#### wishmaster

##### Active member
Hello, wishmaster!

Are you familiar with Long Division?

. . $$\begin{array}{cccccc} &&&& 2 \\ && --&--&-- \\ x^2+1 & | & 2x^2 &-&1 \\ && 2x^2 &+& 2 \\ && --&--&-- \\ &&& - & 3 \end{array}$$

Therefore: .$$\frac{2x^2-1}{x^2+1} \;\;=\;\;2 - \frac{3}{x^2+1}$$

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