Post a New Question

algebra

posted by on .

what are the real or imaginary solutios of the polynomial equation
x^3-8=0

I came up with 1+i sgrt 3 and 1-i sgrt 3
could some see if I'm right

  • algebra - ,

    First byou must find real solution.

    x ^ 3 - 8 = 0 Add 8 to both sides

    x ^ 3 - 8 + 8 = 0 + 8

    x ^ 3 = 8

    x = third root of 8 = 2

    In this case one rational zero is x = 2

    Now you divide polynomial x ^ 3 - 8 with ( x - 2 )

    ( x ^ 3 - 8 ) / ( x - 2 ) = x ^ 2 + 2 x + 4

    OR

    ( x ^ 3 - 8 ) = ( x ^ 2 + 2 x + 4 ) * ( x - 2 ) = 0

    Youe equation have roots when :

    x ^ 2 + 2 x + 4 = 0

    AND

    x - 2 = 0

    Solutions of equation x ^ 2 + 2 x + 4 = 0

    are

    1 + i sgrt 3 and 1 - i sgrt 3


    Soluton of equation x - 2 = 0

    are

    x = 2

    Your equation have 3 solutions.

    One real solution x = 2

    and two imaginary solutions :

    1 + i sgrt 3 and 1 - i sgrt 3

  • algebra - ,

    thank you

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question