Question

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Obtain all the zeroes of the Polynomial p(x) = 2x⁴ - x³ - 11x² + 5x + 5, if two of its zeroes are root 5 and minus root 5.

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Answer 1

Answer :

All the zeroes are : √5, -√5, 1 and -1/2

Solution :

Given zeroes = √5 and -√5

Sum of zeroes = √5 + (-√5)

=> Sum of zeroes = 0

Product of zeroes = √5 (-√5)

=> Product of zeroes = -5

Polynomial = x² - Sx + P

=> Polynomial = x² - 0x + (-5)

=> Polynomial = x² - 5

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Refer to the attachment for division.

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Quotient = 2x² - x - 1

For other zeroes,

=> 2x² - x - 1 = 0

=> 2x² - 2x + x - 1 = 0

=> 2x ( x - 1 ) + 1 ( x - 1 ) = 0

=> ( 2x + 1 )( x - 1 ) = 0

=> 2x + 1 = 0 and x - 1 = 0

=> x = -1/2 and x = 1

•°• All the zeroes are : √5, -√5, 1 and -1/2

Answer 2

$(x {}^{} - \sqrt{5} )(x {}^{} + \sqrt{5)} \: is \: a \: factor$

$\star \: x { }^{2} - 5 \: is \: a \: factor$

By dividing the given polynomial by x² - 5.

Long Division Method:

x² - 5) 2x⁴ - x³ - 11x² + 5x + 5 (2x² - x - 1

          2x⁴       - 10x²

          ---------------------------------

                  - x³   - x² + 5x + 5

                  - x³          + 5x

          ----------------------------------

                             - x²         +  5

  - x²        +  5

----------------------------------

0

         

             

•   Quotient = 2x² - x - 1   

                                             

Now,

$\star \: we \: factorize \: 2x {}^{2} - x - 1$

=> 2x² - x - 1 = 0

=> 2x² - 2x + x - 1 = 0

=> 2x ( x - 1 ) + 1 ( x - 1 ) = 0

=> ( 2x + 1 )( x - 1 ) = 0

=> 2x + 1 = 0 and x - 1 = 0

=> x = -1/2 and x = 1

•°• All the zeroes are : √5, -√5, 1 and -1/2