1. ## Simplify/solve/factor

How do I solve/simplify this:
((2a^-2 b^4) / (4a^3 b^-2))^-2
Do I do the ^-2 to everything inside first? Or what steps do I do first?

And how do I factor this:
x^2-7x+12

I don't remember completely as I haven't done it in years, but it has to add up to -7 and multiply together to 12? Or something along those lines...

Any help would be great, thanks.

2. hi there, please consider the following step to get you started.

$\left(\frac{2a^{-2} b^4}{4a^3 b^{-2}}\right)^{-2} = \left(\frac{4a^3 b^{-2}}{2a^{-2} b^4}\right)^{2}$

Now square every term inside the bracket then simplify.

3. Originally Posted by alfer22
H

And how do I factor this:
x^2-7x+12

I don't remember completely as I haven't done it in years, but it has to add up to -7 and multiply together to 12? Or something along those lines...

yes, a and b are factors of 12 that give you a+b = -7

factored form will be $(x+a)(a+b)$

Spoiler:
$x^2-7x+12 = (x-3)(x-4)$

4. Originally Posted by pickslides
hi there, please consider the following step to get you started.

$\left(\frac{2a^(-2) b^4}{4a^3 b^{-2}}\right)^{-2} = \left(\frac{4a^3 b^{-2}}{2a^{-2} b^4}\right)^{2}$

Now square every term inside the bracket then simplify.
So whenever something is bracketed by a negative exponent you flip the inside and change the exponent to positive?

5. correct.

6. $\left(\frac{8y^6 z^{\left(\frac{2}{3}\right)}}{x\left(\frac{1}{3}\r ight)}\right)^{3}$

I am clueless on how to deal with the fraction exponents.

The final answer should come to $512xy^{18}z^{2}$, but I do not know how to get there.

Any help would be great.

7. I think you mean?

$\left(\frac{8y^6 z^{\frac{2}{3}}}{x^\frac{1}{3}}\right)^{3}
$

If so multiply the 3 into each term

$\frac{8^3y^{6\times 3} z^{\frac{2}{3}\times 3}}{x^{\frac{1}{3}\times 3}}$

$\frac{512y^{18} z^{2}}{x^{1}}$

$512x^{-1}y^{18} z^{2}$

8. If I have something like:
$(x^{3}y^{-\left(\frac{1}{2}\right)})^{-2}$

How do I deal with the negative exponent then?

9. Remember these rules

$a^{-n} = \frac{1}{a^n}$

and

$\frac{1}{a^{-n}}=a^{n}$