Post a New Question

Precalculus

posted by on .

x^4-4x^3+3x^2+8x-16

Find the zeroes given one is x^2-x-4.

I don't know what to do now. i tried synthetic division. pls help : D

  • Precalculus - ,

    x^4-4x^3+3x^2+8x-16
    --------------------
    x^2 - x - 4

    = x^2 - 3 x + 4 by long division

    x = [3 +/- sqrt (9-16) ]/2

    [3 +/- i sqrt (7)]/2
    for the second pair of complex roots

    for the original 2 roots
    x = [ 1 +/- sqrt(1+16)]/2

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question