Can somebody tell me what to do next for solving polynomial equations?
x^3 - 3x^2 -4x + 12= 0
I got the factors of 12 and i used 2.
(x+2) (x^2 + x + 12)=0
What do I do next?
I used synthetic division to get (x+2)(x^2 + x + 12)=0
Factor x^2+x+12 now
(x+4)(x+3)=0
therefore, (x+2)(x+4)(x+3)=0?
x= -2, -4, -3?
(x+3)(x+5)
To solve the polynomial equation (x^3 - 3x^2 - 4x + 12 = 0) after factoring, you first set each factor equal to zero and solve for x.
Setting (x + 2) equal to zero:
x + 2 = 0
x = -2
Next, you need to solve the remaining quadratic equation (x^2 + x + 12 = 0). However, upon closer inspection, you'll see that this quadratic equation does not have real solutions since the discriminant (b^2 - 4ac) is negative.
Therefore, the only real solution for the original polynomial equation is x = -2.