1. simplified form of x^4-81/x+3

2. simplified form of (x^2yz)^2(xy^2z^2)/(xyz)^2

Question

1. simplified form of x^4-81/x+3

2. simplified form of (x^2yz)^2(xy^2z^2)/(xyz)^2

lgbasallote · Answer

1) hmm if you notice you have difference of two squares there... (x^2)^2 - (9)^2 yes? try factoring that out..

anonymous · Answer

$$\Large \frac{(x^2yz)^2(xy^2z^2)}{(xyz)^2}$$

Use the rule.. $$\LARGE \frac{a^m}{a^n}=a^{m-n}$$

And you'll have...

$$\LARGE y^2z^2 x^3$$

anonymous · Answer

wait im confused about the first one

anonymous · Answer

so would the first one be (x^2 +9)(x^2-9)??

anonymous · Answer

@haileystowers so you don't understand the first question??

anonymous · Answer

no

anonymous · Answer

$$\LARGE x^4-\frac{81}{x}+3$$ .. hmmm  actually , I don't know what to simplify there, do you have multiple choices ?? :S

anonymous · Answer

@lgbasallote  can you explain please how do you simplify this one, It seems I don't get it :( ...  (first question ! )

lgbasallote · Answer

it's (x^4 - 81)/(x+3)

=)))

now..doesnt that make more sense /:)

lgbasallote · Answer

you finish it @Kreshnik ^_^

anonymous · Answer

auff... I thought it was the one I wrote a few minutes ago  LOL ... thnx... here you go ...

$$\LARGE \frac{x^4-81}{x+3}=\frac{(x^2)^2-9^2}{x+3}$$  now use this formula...
$$\LARGE a^2-b^2=(a-b)(a+b)$$

$$\LARGE \frac{(x^2-9)(x^2+9)}{x+3}$$  Final Answer !!