Question 32

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Jambmaths question: 

Find the roots of ${{x}^{3}}-2{{x}^{2}}-5x+6=0$

Option A: 

1, 2, –3

Option B: 

–1, –2, 3

Option C: 

–1, 2, –3

Option D: 

1, –2, 3

Jamb Maths Solution: 

\[\begin{align}  & \text{Let }f(x)={{x}^{3}}-2{{x}^{2}}-5x+6,\text{ using trial and error method, let }x=1 \\ & f(1)={{1}^{3}}-2{{(1)}^{2}}-5(1)+6=0 \\ & \text{Since }f(1)=0,\text{ }x-1\text{ is a factor of }f(x) \\ & \text{Using long division} \\ & x-1\overset{{{x}^{2}}-x-6}{\overline{\left){\begin{align}  & {{x}^{3}}-2{{x}^{2}}-5x+6 \\ & \underline{{{x}^{3}}-{{x}^{2}}\text{              }} \\ & \text{      }{{x}^{2}}-5x+6 \\ & \text{     }\underline{{{x}^{2}}+x}\text{    } \\ & \text{          }-6x+6 \\ & \text{         }\underline{\text{  }-6x+6} \\ & \text{           }------ \\\end{align}}\right.}} \\ & f(x)=(x-1)({{x}^{2}}-x-6)=(x-1)(x-3)(x+2) \\ & \text{The roots of }f(x)\text{ are }x=1,\text{ }x=-2,\text{ }x=3 \\\end{align}\]

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