I do not know how to solve this question. 5x^{3}+30x=0
Hi LaTanya;
5x^{3}+30x=0
For the FOIL, we know that...
FIRST must be (5x^{2})(x) or (x^{2})(5x)
LAST must include 0, (?)(0) or (0)(??)
There are four ways to factor this.
(5x^{2}+30)(x+0)=0
FOIL...
FIRST...(5x^{2})(x)=5x^{3}
OUTER...(5x^{2})(0)=0
INNER...(30)(x)=30x
LAST...(30)(0)=0
Factored...
(5x^{2}+30)(x+0)=0
Either parenthetical equation can be equal to zero...
5x^{2}+30=0 or (x+0)=0
5x^{2}=-30 or x=0
x^{2}=-30/5
x^{2}=-6
The first parenthetical equation does NOT work. There is no such thing as the square root of a negative number.
However, one possibility is...
x=0
****************
Let's factor again...
(x^{2}+6)(5x+0)=0
FOIL...
FIRST...(x^{2})(5x)=5x^{3}
OUTER...(x^{2})(0)=0
INNER...(6)(5x)=30x
LAST...(6)(0)=0
(x^{2}+6)(5x+0)=0
Either parenthetical equation can be =0.
x^{2}+6=0 or 5x+0=0
x^{2}=-6 or 5x=0, x=0
This also produced the only result of x=0 because there is no such thing as a the square root of a negative number.
x=0
***************
Let's factor again...
(-x^{2}-6)(-5x+0)=0
FIRST...(-x^{2})(-5x)=5x^{3}^{
}
OUTER...(-x^{2})(0)=0
INNER...(-6)(-5x)=30x
LAST...(-6)(0)=0
(-x^{2}-6)(-5x+0)=0
-x^{2}-6=0 or -5x+0=0
-x^{2}=6 or x=0
x=0 is the only solution.
******************
Let's factor again...
5x^{3}+30x=0
5x(x^{2}+6)=0
x=0, to produce 5x as 0, is one solution.
x^{2}+6=0
x^{2}=-6 cannot be resolved to a positive number.
The only solution is x=0