I have ten algebra questions but I'm having a hard time answering these two, please help?

Question

1. Solve the equation, noting any double roots.
x^3+9x=6x^2


2. Solve by factoring. 6x^3+3x^2-9x=0
Why does it equal to 0?

Thanks everyone. :0)
You all helped me understand it.

Merry Christmas to all!

Answer 1

Question 1
x (x² - 6x + 9) = 0
x (x - 3) (x - 3) = 0
x = 0 , x = 3

Question 2
3x (2x² + x - 3) = 0
3x (2x + 3) (x - 1) = 0
x = 0 , x = - 3/2 , x = 1

Answer 2

For 1, first subtract 6x^2, then factor out x, then factor again.

x^3 +9x = 6x^2
x^3 - 6x^2 +9 x = 0
x(x^2 -6 + 9) = 0
x (x-3) (x-3) = 0

therefore x = 0 or +3.

For 2,
you are given:

6x^3+3x^2-9x=0
3x (2x^2 + 1 x - 3) = 0
3x (2x + 3) (x - 1) = 0

Therefore x = 0, -1.5, or 1

Notice how factoring involves certain guessing and checking.

Answer 3

1. set up so equal to zero so x^3+9x-6x^2=0
now solve so x(x^2-6x+9)=0
so (x)(x-3)(x-3)=0 so the roots are 0 and 3 with 3 being a double root
2.same thing here so 3x(2x+3)(x-1) so x=0, (-3/2), and 1
because the equation needs to be set to zero in order to solve it.

Answer 4

1.
x^3 + 9x = 6x^2
x^3 - 6x^2 + 9x = 0
x ( x^2 - 6x +9 ) = 0
x ( x - 3 ) ( x - 3 ) = 0
Thus,
x = 0
or
x = 3

2.
6x^3 + 3x^2 - 9x = 0
3x ( 2x^2 + x - 3 ) = 0
3x ( x - 1 ) ( 2x + 3 ) = 0
Thus,
3x = 0 which becomes x = 0
or
x = 1
or
2x = -3 which becomes x = -3/2

Please pick me as best answer :D I'm a fellow student like you. I'm doing my homework now which involves questions like these. Hahaha.

Answer 5

1 Ans: x=3,x=3&x=0;
2.Ans: x=1,x=0 & x=3/2

First synthetically divide those two cubic eqns., then they'll become quadratic eqns which can be solved using
[-b(+-)sqrt(b^2-(4*a*c))] / 2*a....