Question

Find the polynomial whose zeros are cubes of the zeros of polynomial 4x²- 17x -21.​

Answer 1

Answer:

p(x) = 4x² - 17x - 21

      α + β = -b/a = 17/4 -------------------- (i)

         αβ = c/a = -21/4 ---------------------(ii)

Given that the roots of new polynomial is cube of the roots of existing polynomial :

      i.e. α³ and β³

Now ,

  the polynomial is

              x² - ( α³ + β³ )x + α³β³ = 0   ( General form by roots )

Now, α³ + β³ = ( α + β )³ - 3αβ ( α + β )

                     = ( 17/4 )³ - 3( -21/4 )( 17/4 )        [ from eq(i) ]

                     = 4913/64 + ( 63/4 )( 17/4 )

                     = 4913/64 + 1071/16

                     = 9197/64

and, α³β³ = ( αβ )³ = ( -21/4 )³ = -9261/64   [ from eq(ii) ]

Forming the polynomial,

        x² - ( 9197/64 )x + ( -9261/64 ) = 0

    or 64x² - 9197x - 9261 = 0          (Ans.)

       Thank U

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