MHB Solve for Rational Term: 2x^2-1 Over x^2+1

[theakdad]

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

 

[I like Serena]

U have a question why is $$\frac{2x^2-1}{x^2+1}$$ equal to $$2-\frac{3}{x^2+1}$$ ??

\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}

 

[theakdad]

\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?

 

[I like Serena]

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, $$\frac 7 2 = 3 + \frac 1 2$$.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]

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}<br /> &amp;&amp;&amp;&amp; 2 \\<br /> &amp;&amp; --&amp;--&amp;-- \\<br /> x^2+1 &amp; | &amp; 2x^2 &amp;-&amp;1 \\<br /> &amp;&amp; 2x^2 &amp;+&amp; 2 \\<br /> &amp;&amp; --&amp;--&amp;-- \\<br /> &amp;&amp;&amp; - &amp; 3 \end{array}Therefore: .\frac{2x^2-1}{x^2+1} \;\;=\;\;2 - \frac{3}{x^2+1}

 

[theakdad]

Hello, wishmaster!


Are you familiar with Long Division?

. . \begin{array}{cccccc}<br /> &amp;&amp;&amp;&amp; 2 \\<br /> &amp;&amp; --&amp;--&amp;-- \\<br /> x^2+1 &amp; | &amp; 2x^2 &amp;-&amp;1 \\<br /> &amp;&amp; 2x^2 &amp;+&amp; 2 \\<br /> &amp;&amp; --&amp;--&amp;-- \\<br /> &amp;&amp;&amp; - &amp; 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!