Quick Algebra Questions: Factoring and Solving Equations Explained

[physicszman]

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

 

[ShawnD]

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.

 

[himanshu121]

U use calulators for these pros

U can write $$y^2-2=y^2-\sqrt(2)^2$$

 

[Kacper]

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.