# Thread: Simplifying and finding the domain

1. ## Simplifying and finding the domain

So i transferred schools and i was in algebra 2 last year, now im in precal and the school is a lot more difficult and there are some things i havent covered.

Can u give me the steps on how u did it?because if i dont learn anything itll be useless

how do i find the domain of 4x^3+3 when x greater than or equal to 0, and the domain of x+1 over 2x+1

also how do i simplify this? 4(3-x) / x-3

and another x+13/x^3(3-x) * x(x-3)/5

can i simplify this? x^2-36/6-x im at (x+6)(x+6)/6-x can i simplify it any further?

last one: simplify the complex fraction x-4 / x/4-4/x (writing it down on paper would make it more clear with the / as fraction)
-thanks

2. You made need to use some brackets so these problems make a bit more sense.

$\displaystyle \frac{x^2-36}{6-x}=\frac{(x+6)(x-6)}{6-x}\neq \frac{(x+6)(x+6)}{6-x}$

3. Simplify: 4(3-x) / x-3
(I don't know if this is correct or not)

Suppose that it is an equation, with an equals:

$y=\dfrac{4(3-x)}{x-3}$

$-1 * y=-1 * \dfrac{4(3-x)}{x-3}$

$-1y=\dfrac{4(x-3)}{x-3}$

$-1y=4$

$y=-4$

$\dfrac{4(3-x)}{x-3} = -4, x\ne{3}$

It has lost some data when it is simplified to -4, because x cannot be 3.

Is the second simplify question:

$\dfrac{x+13}{x^3(3-x)} * \dfrac{x(x-3)}{5}$ ?

And the forth one:

$\dfrac{x-4}{\frac{x}{4}-\frac{4}{x}}$ ?

4. $(3-x) = -(x-3)$

$\Rightarrow \dfrac{4(3-x)}{x-3} = \dfrac{-4(x-3)}{x-3} = -4.$

how do i find the domain of 4x^3+3 when x greater than or equal to 0, and the domain of x+1 over 2x+1
I find this confusing. Do you mean the range?

5. Originally Posted by TheCoffeeMachine
$(3-x) = -(x-3)$

$\Rightarrow \dfrac{4(3-x)}{x-3} = \dfrac{-4(x-3)}{x-3} = -4.$

I find this confusing. Do you mean the range?
my friend said the problem was already done, so maybe its the books problem.

Originally Posted by Educated
Simplify: 4(3-x) / x-3
(I don't know if this is correct or not)

Suppose that it is an equation, with an equals:

$y=\dfrac{4(3-x)}{x-3}$

$-1 * y=-1 * \dfrac{4(3-x)}{x-3}$

$-1y=\dfrac{4(x-3)}{x-3}$

$-1y=4$

$y=-4$

$\dfrac{4(3-x)}{x-3} = -4, x\ne{3}$

It has lost some data when it is simplified to -4, because x cannot be 3.

Is the second simplify question:

$\dfrac{x+13}{x^3(3-x)} * \dfrac{x(x-3)}{5}$ ?

And the forth one:

$\dfrac{x-4}{\frac{x}{4}-\frac{4}{x}}$ ?
yes for the second, and yes to the 4th