Quick Algebra Questions: Factoring and Solving Equations Explained

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The discussion addresses several algebra problems, focusing on factoring and solving equations. For the equation 6x - x^2 = x, the correct factoring leads to solutions x = 0 and x = 5, which were confirmed with a calculator. The second equation, 5 - 35/x^2 = 0, simplifies to x = ±√7. The factoring of 2x^4 - 32 was corrected to 2(x^2 + 4)(x + 2)(x - 2), while the factorization of 5y^3 + 2y^2 - 10y - 4 was discussed with a focus on grouping. Overall, the thread provides clarifications and corrections for algebraic misconceptions.
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Hi, this is a bit of topic but I could not find any good algebra forums. Havn't done this since freshman yr so I am a bit rusty.

1) 6x - x^2 = x

*I set this to zero and factor and get -2, and 3. But it doesn't work when i plug it back in? where did I mess up?

2) 5 - 35/x^2 = 0

*No idea how to do it?

3) Factor : 2x^4 - 32
2(x+2)(x-2)(x+2)(x-2)

4) Factor : 5y^3 + 2y^2 - 10y - 4
(5y^3 + 2y^2)(-10y - 4)
y^2(5y + 2) -2(5y + 2)
(y^2 - 2)(5y + 2)

* I think i can factor it more i don't know how.

any help is appreciated
 
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Originally posted by physicszman
1) 6x - x^2 = x

-x^2 + 5x = 0
x(-x + 5) = 0

x = 0
x = 5

My calculator confirmed these answers.



2) 5 - 35/x^2 = 0

-35/x^2 = -5
-35/-5 = x^2
x^2 = 7
x = sqrt(7)
x = -sqrt(7)

My calculator confirmed these answers.

3) Factor : 2x^4 - 32
2(x+2)(x-2)(x+2)(x-2)
Your answer doesn't work. If you multiply those terms together you do not get what you started with.

2(x^4 - 16)
2(x^2 + 4)(x^2 - 4)
2(x^2 + 4)(x + 2)(x - 2)

4) Factor : 5y^3 + 2y^2 - 10y - 4
(5y^3 + 2y^2)(-10y - 4)
y^2(5y + 2) -2(5y + 2)
(y^2 - 2)(5y + 2)

I don't remember too much about grouping but it looks pretty good.
 
U use calulators for these pros

U can write y^2-2=y^2-\sqrt(2)^2
 


Your answer doesn't work. If you multiply those terms together you do not get what you started with.

2(x^4 - 16)
2(x^2 + 4)(x^2 - 4)
2(x^2 + 4)(x + 2)(x - 2)

It is done correctly the answer is just extraneous.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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