+26402 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 \(\mathbf{x_1=3 \qquad x_2 = -4 \qquad x_3= -\frac{1}{2}}\)
+26402 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 \(\mathbf{x_1=3 \qquad x_2 = -4 \qquad x_3= -\frac{1}{2}}\)
