I can't remember how to do this. Can someone please help me refresh my memory? I have to simplify these expressions.
1. x^3-8
x-2
2. 1 - 1
x + h x
3. 2
x^2
10
x^5
4. 2x - 1 - 8
x^2-6x+9 x+1 x^2-2x-3
Hello, journeygirl280!
I think I deciphered the last two problems . . .
We have: .$\displaystyle \frac{2}{x^2} \div \frac{10}{x^5} \;= \;\frac{2}{x^2}\cdot\frac{x^5}{10} \;=\;\frac{x^3}{5}$$\displaystyle 3)\;\;\frac{\frac{2}{x^2}}{\frac{10}{x^5}} $
We have: .$\displaystyle \frac{2x}{(x-3)^2} - \frac{1}{x+1} - \frac{8}{(x+1)(x-3)} $$\displaystyle 4)\;\;\frac{2x}{x^2-6x+9} - \frac{1}{x+1} - \frac{8}{x^2-2x-3}$
Get a common denominator:
. . $\displaystyle \frac{2x}{(x-3)^2}\cdot{\color{blue}\frac{x+1}{x+1}} \,-\,\frac{1}{x+1}\cdot{\color{blue}\frac{(x-3)^2}{(x-3)^2}} \,-\,\frac{8}{(x+1)(x-3)}\cdot{\color{blue}\frac{x-3}{x-3}} $
. . $\displaystyle = \;\frac{2x(x+1)-(x-3)^2 - 8(x-3)}{(x-3)^2(x+1)} \;= \;\frac{2x^2 + 2x - x^2 + 6x - 9 - 8x + 24}{(x-3)^2(x+1)}
$
. . $\displaystyle = \;\frac{x^2+15}{(x-3)^2(x+1)} $