If  \alpha ,\beta ,\gamma are roots of 2x^{3}-x^{2}-1= 0 then \left ( 2-\alpha \right )\left ( 2-\beta \right )\left ( 2-\gamma \right ) equals

  • Option 1)

    10

  • Option 2)

    \frac{11}{2}

  • Option 3)

    12

  • Option 4)

    \frac{13}{2}

 

Answers (1)
A Aadil Khan

Using factor theorem : 2x^{3}-x^{2}-1=2\left ( x-\alpha \right )\left ( x-\beta \right )\left ( x-\gamma \right )

Now putting x=2 both sides we get 

\left ( 2-\alpha \right )\left ( 2-\beta \right )\left ( 2-\gamma \right )=\frac{11}{2}

\therefore Option (B)

 

Factor Theorem -

Any polynomial can be written in terms of product of its factors.

- wherein

If P\left ( x \right )= 0  has roots  \alpha _{1},\alpha _{2},\cdots \alpha _{n}  and  P\left ( x \right )  has leading coefficient 'a' then then P\left ( x \right )= a\left ( x-\alpha _{1} \right )\left ( x-\alpha _{2} \right )\cdots \left ( x-\alpha _{n} \right )  

 

 


Option 1)

10

This is incorrect

Option 2)

\frac{11}{2}

This is correct

Option 3)

12

This is incorrect

Option 4)

\frac{13}{2}

This is incorrect

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