SOLUTION: <pre> Solving equations: find the real number solutions of the equation. g³ + 3g² - g - 3 = 0 </pre> any help would be appreciated, thank you, Mike

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: <pre> Solving equations: find the real number solutions of the equation. g³ + 3g² - g - 3 = 0 </pre> any help would be appreciated, thank you, Mike      Log On


   



Question 564891:
Solving equations: find the real number solutions of the equation.

     g³ + 3g² - g - 3 = 0

any help would be appreciated, thank you, Mike

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
     g³ + 3g² - g - 3 = 0

Factor the first two terms g³ + 3g² by taking out the common factor 
g², getting g²(g + 3).
Factor the last two terms -g - 3 by taking out the common factor 
-1, getting -1(g + 3).
So the equation is now:

 g²(g + 3) - 1(g + 3) = 0

Now factor out the common factor (g + 3)

      (g + 3)(g² - 1) = 0

Now the expression in the second parentheses g² - 1 can be factored as the
difference of two squares as (g - 1)(g + 1) and the equation is now:

(g + 3)(g - 1)(g + 1) = 0

Us the principle of zero factors and set each factor = 0, and solve:

   g + 3 = 0;   g - 1 = 0;   g + 1 = 0
       g = -3;      g = 1;       g = -1

The solutions are these three: -3, 1, and -1.

Edwin