# Thread: Subtraction of fractions with polynomials

1. ## Subtraction of fractions with polynomials

For a problem I'm working on, I get:
$\displaystyle \frac{1}{x}-\frac{2x}{x^{2}+1}$

My book has:
$\displaystyle \frac{1-x^{2}}{x(x^{2}+1)}$

Is this the same?

2. Yes, you can verify by cross multiplying

$\displaystyle \dfrac{1}{x} \cdot \dfrac{x^2+1}{x^2+1} = \dfrac{x^2+1}{x(x^2+1)}$

$\displaystyle \dfrac{2x}{x^2+1} \cdot \dfrac{x}{x} = \dfrac{2x^2}{x(x^2+1)}$

Now the original expression can be rewritten as $\displaystyle \dfrac{x^2+1}{x(x^2+1)} - \dfrac{2x^2}{x(x^2+1)}$

Can you finish?

3. Originally Posted by e^(i*pi)
Yes, you can verify by cross multiplying

$\displaystyle \dfrac{1}{x} \cdot \dfrac{x^2+1}{x^2+1} = \dfrac{x^2+1}{x(x^2+1)}$

$\displaystyle \dfrac{2x}{x^2+1} \cdot \dfrac{x}{x} = \dfrac{2x^2}{x(x^2+1)}$

Now the original expression can be rewritten as $\displaystyle \dfrac{x^2+1}{x(x^2+1)} - \dfrac{2x^2}{x(x^2+1)}$

Can you finish?
$\displaystyle \dfrac{x^2+1}{x(x^2+1)} - \dfrac{2x^2}{x(x^2+1)} =$

$\displaystyle \dfrac{x^2+1-2x^2}{x(x^2+1)} =$

$\displaystyle \dfrac{1-x^2}{x(x^2+1)}$