# 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) $\frac{13}{2}$

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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