# Is there a faster way to simplify this fraction?

#### gudin727

I succed in simplifing this fraction, but I did this in the long way- using ax+b and mx+n.
I am pretty sure there is a shorter way... what do you think?

#### skeeter

MHF Helper
faster way?

$$\displaystyle \frac{x^2+10}{(x^2+4)(x^2+1)} = \frac{-2}{x^2+4} + \frac{3}{x^2+1}$$

$$\displaystyle \frac{x^2+10}{(x^2+4)(x^2+1)} = \frac{-2(x^2+1)}{(x^2+4)(x^2+1)} + \frac{3(x^2+4)}{(x^2+4)(x^2+1)}$$

numerator forms the equation ...

$$\displaystyle x^2+10 = -2(x^2+1) + 3(x^2+4)$$

$$\displaystyle 10 = 10$$

solution set ... $x \in \mathbb{R}$

#### Prove It

MHF Helper
Am I wrong in thinking that

\displaystyle \begin{align*} \frac{x^2 + 10}{\left( x^2 + 4 \right) \left( x^2 + 1 \right) } &= \frac{x^2 + 4 + 6}{\left( x^2 + 4 \right) \left( x^2 + 1 \right) } \\ &= 1 + \frac{6}{x^2 + 1} \end{align*}

is much simpler?

#### skeeter

MHF Helper
Am I wrong in thinking that

\displaystyle \begin{align*} \frac{x^2 + 10}{\left( x^2 + 4 \right) \left( x^2 + 1 \right) } &= \frac{x^2 + 4 + 6}{\left( x^2 + 4 \right) \left( x^2 + 1 \right) } \\ &= 1 + \frac{6}{x^2 + 1} \end{align*}

is much simpler?
does that work if x = 0 ?

#### Plato

MHF Helper
Am I wrong in thinking that
\displaystyle \begin{align*} \frac{x^2 + 10}{\left( x^2 + 4 \right) \left( x^2 + 1 \right) } &= \frac{x^2 + 4 + 6}{\left( x^2 + 4 \right) \left( x^2 + 1 \right) } \\ &= 1 + \frac{6}{x^2 + 1} \end{align*}
is much simpler?
Well $\dfrac{x^2 + 4 + 6}{\left( x^2 + 4 \right)(x^2+1)}\ne\dfrac{x^2 + 4 }{\left( x^2 + 4 \right)}+\dfrac{ 6}{\left( x^2 + 1 \right)}$

#### Prove It

MHF Helper
No it doesn't, I was half asleep when I wrote that hahaha.

#### skeeter

MHF Helper
No it doesn't, I was half asleep when I wrote that hahaha.
It happens to all of us ... no worries.