- Thread starter
- #1

#### 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:

- Thread starter wishmaster
- Start date

- Thread starter
- #1

- Oct 11, 2013

- 211

Last edited:

- Admin
- #2

- Mar 5, 2012

- 9,152

\begin{aligned}U have a question why is \(\displaystyle \frac{2x^2-1}{x^2+1}\) equal to \(\displaystyle 2-\frac{3}{x^2+1}\) ??

\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}

- Thread starter
- #3

- Oct 11, 2013

- 211

thank you!\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}

Is this the only way to caclulate it?

- Admin
- #4

- Mar 5, 2012

- 9,152

It is similar to calculating:thank you!

Is this the only way to caclulate it?

\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.

[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]

- Thread starter
- #6

- 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!