Can not get this.

Question

2x^3+3x^2-2x-3=0   I can not get this. Please explain

Answer 1

I assume this problem is from the Algebra, because in Precalculus there is another method we might use. Third power equation has no more than three roots in set of real numbers. To solve this kind of equation we have to factorize it ax³ + bx² + cx + d = a(x - x₁)(x - x₂)(x - x₃) The procedure is to figure one of them out first. For given equation x₁ = 1 , let's divide given equation by (x-1) :
             2x² + 5x + 3                                                                                       
   (x-1) l (2x³ + 3x² - 2x - 3)     
        -   (2x³ - 2x² )      
                     5x²   - 2x       
               -   (5x²   - 5x)         
                               3x  - 3  
                           -  (3x - 3)              
                                   0           
----> 2x³ + 3x² - 2x - 3 = (x-1)(2x²+5x+3), now we have to factorize 2x² + 5x + 3 = (x + 1)(2x + 3)
        2x³ + 3x² - 2x - 3 = (x - 1)(x + 1)(2x + 3) and roots are {-1, -3/2, 1}

Answer 2

You have some terms with similar coefficients: 2x³  and -2x, 3x² - 3. So we can try factoring by grouping: 

2x(x² - 1) + 3 (x² - 1) = 0

(2x + 3)(x² - 1) = 0

The second parentheses has the difference of two squares, so We can continue to factor:

(2x + 3)(x + 1)(x - 1) = 0

Since the product equals 0, then one of the factors must be 0, therefore 
2x + 3 = 0   leads to x = -3/2

x + 1 = 0  leads to x = -1

x -1 = 0 leads to x = 1

So x = -3/2, -1, 1