Find the value of x where the curve $y={{x}^{3}}+2{{x}^{2}}-5x-6$cross the x – axis
–2, 1 and –3
2, –1 and –3
2, 1 and 3
–2, –1and 3
$\begin{align} & {{x}^{3}}+2{{x}^{2}}-5x-6=0 \\ & f(x)={{x}^{3}}+2{{x}^{2}}-5x-6 \\ & \text{Let }x=\text{1} \\ & f(-1)={{(-1)}^{3}}+2{{(-1)}^{2}}-5(-1)-6 \\ & f(-1)=-1+2+5-6=0 \\ & f(-1)=0,\text{ (}x+1)\text{ is a factor of }f(x) \\ & \text{Using long division} \\ & x+1\overset{{{x}^{2}}+x-6}{\overline{\left){\begin{align} & {{x}^{3}}+2{{x}^{2}}-5x-6 \\ & \underline{{{x}^{3}}+{{x}^{2}}} \\ & \text{ }{{x}^{2}}-5x-6 \\ & \text{ }\underline{\text{ }{{x}^{2}}+x} \\ & \text{ }{{x}^{2}}-5x-6 \\ & \text{ }\underline{\text{ }{{x}^{2}}+x\text{ }} \\ & \text{ }-6x-6 \\ & \text{ }\underline{\text{ }-6x-6} \\ & \text{ }------ \\\end{align}}\right.}} \\ & f(x)=(x+1)({{x}^{2}}+x-6) \\ & f(x)=(x+1)(x+3)(x-2) \\ & (x+1)(x+3)(x-2)=0 \\ & x=-1.-3\text{ and }2 \\\end{align}$