MATH

Find all zeros of f(x)=2x^3+3x^2-23x-12

$2x^3+3x^2-23x-12=0$

1. Try all divider of 12: $\pm 1, ~ \pm 2,~ \pm 3,~ \pm4, ~\pm 6,~\pm12$

We find $x_1 = 3$ and $x_2 = -4$

2. 

$(x-3)(x+4) = x^2+4x-3x-12 = x^2+x-12$

$2x^3+3x^2-23x-12 : x^2+x-12=2x+1$

$2x_3+1 = 0\\ 2x_3 = -1\\ x_3 = -\frac{1}{2}$

We have zeros  $\mathbf{x_1=3 \qquad x_2 = -4 \qquad x_3= -\frac{1}{2}}$