1. Simplifying expressions

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

2. Originally Posted by journeygirl280
1. x^3-8
x-2
Hint: Difference two two cubes in numerator.

2. 1 - 1
x + h x
With common denominator,
$\frac{x}{x(x+h)} - \frac{x+h}{x(x+h)}$.
Now continue.

The last two do not make any sense.

3. Hello, journeygirl280!

I think I deciphered the last two problems . . .

$3)\;\;\frac{\frac{2}{x^2}}{\frac{10}{x^5}}$
We have: . $\frac{2}{x^2} \div \frac{10}{x^5} \;= \;\frac{2}{x^2}\cdot\frac{x^5}{10} \;=\;\frac{x^3}{5}$

$4)\;\;\frac{2x}{x^2-6x+9} - \frac{1}{x+1} - \frac{8}{x^2-2x-3}$
We have: . $\frac{2x}{(x-3)^2} - \frac{1}{x+1} - \frac{8}{(x+1)(x-3)}$

Get a common denominator:

. . $\frac{2x}{(x-3)^2}\cdot{\color{blue}\frac{x+1}{x+1}} \,-\,\frac{1}{x+1}\cdot{\color{blue}\frac{(x-3)^2}{(x-3)^2}} \,-\,\frac{8}{(x+1)(x-3)}\cdot{\color{blue}\frac{x-3}{x-3}}$

. . $= \;\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)}
$

. . $= \;\frac{x^2+15}{(x-3)^2(x+1)}$

4. Originally Posted by journeygirl280
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
Different hint: $x^3-8=0$ when $x=2$ so $x-2$ is a factor of $x^3-8$.

(well really the same hint in disguise)

RonL