Question 31

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Jambmaths question: 

Find the value of x where the curve $y={{x}^{3}}+2{{x}^{2}}-5x-6$cross the x – axis

Option A: 

–2, 1 and  –3

Option B: 

2, –1 and –3

Option C: 

2, 1 and 3

Option D: 

–2, –1and 3

Jamb Maths Solution: 

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

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