find solutions to the equation?

Question

(x+2)((x^3)+(2x^2)-3x)=0 if no solutions exists then explain why

I have no idea even where to begin

Answer 1

To solve, find each value of x that will cause the equation to equal zero. To do so, break the equation into chunks.

Equation: (x+2)((x^3)+(2x^2)-3x)=0

1.) Rule: 0 * x = 0: therefore, you can split this into 2 parts:
a.) (x+2) = 0
b.) (x^3 + 2x^2 - 3x) = 0

a.) x + 2 = 0
x = -2 (Solution #1)

b.) (x^3 + 2x^2 - 3x) = 0
To solve polynomials, you need to break it down into an easier form of (x+a)* (x+b) * (x+c)...

First, all 3 parts of this equation are divisible by x, so you can rewrite it:
(x) * (x^2 + 2x - 3) = 0

Using the rule above, you can split the equation again:
x = 0 (Solution #2)

c.) (x^2 + 2x - 3) = 0
When a binomial (like what we have above) has a prime number at the end, it's pretty easy. Since the only way to multiply 2 integers and get 3 is (3 * 1), and the number is negative, you know that the solution is either:
c1.) (x+3) * (x-1) = 0 or
c2.) (x-3) * (x+1) = 0

Since you have +2x in the middle instead of -2x, then it must be c1.
c1.) (x+3) * (x-1) = 0
Anything * 0 = 0, so:
x+3 = 0
x = -3 (Solution #3)
x -1 = 0
x = 1 (Solution #4).

Final: x = -2, 0 , -3, 1
Check:
1.) (-2+2)((-2^3)+(2(-2)^2)-3(-2))=0
0 * (-8 + 8 +6) = 0
0 * 6 = 0 (correct)

2.) (0+2)((0^3)+(2(0)^2)-3(0))=0
2 * (0 + 0 + 0) = 0
2 * 0 = 0 (correct)

3.) (-3+2)((-3^3)+(2(-3)^2)-3(-3)=0
-1 * (-27 + 18 +9)
-1 * 0 = 0 (correct)

4.) (1+2)((1^3)+(2(1)^2-3(1))=0
3 * (1 + 2 - 3) = 0
3 * 0 = 0 (correct)