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  • I have a doubt, kindly clarify. - Complex numbers and quadratic equations - JEE Main

If  \alpha ,\beta ,\gamma ,\delta  are roots of  x^{4}+x^{3}+x^{2}+1= 0  then \left ( 1+\alpha \right )\left ( 1+\beta \right )\left ( 1+\gamma \right )\left ( 1+\delta \right )  equals

  • Option 1)

    -1

  • Option 2)

    0

  • Option 3)

    1

  • Option 4)

    2

 

Answers (1)

best_answer

\left ( 1+\alpha \right )\left ( 1+\beta \right )\left ( 1+\gamma \right )\left ( 1+\delta \right )=\left ( -1-\alpha \right )\left ( -1-\beta \right )\left ( -1-\gamma \right )\left ( -1-\delta \right )

Now, x^{4}+x^{3}+x^{2}+1=1\left ( x-\alpha \right )\left ( x-\beta \right )\left ( x-\gamma \right )\left ( x-\delta \right )

Putting x=-1 both sides -

1-1+1+1=\left ( -1-\alpha \right )\left ( -1-\beta \right )\left ( -1-\gamma \right )\left ( -1-\delta \right )

\Rightarrow \left ( 1+\alpha \right )\left ( 1+\beta \right )\left ( 1+\gamma \right )\left ( 1+\delta \right )=2

\therefore Option (D)

 

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)

-1

This is incorrect

Option 2)

0

This is incorrect

Option 3)

1

This is incorrect

Option 4)

2

This is correct

Posted by

divya.saini

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