Final answer:
The question deals with finding the remaining zeroes of the polynomial x⁴-4x²-4x-1 given two zeroes. Two other zeroes are complex and are conjugates since the polynomial has real coefficients. Multiplying the given factors will provide a quadratic factor which can be used to find the other two zeroes.
Step-by-step explanation:
You are asked to obtain all other zeroes of the polynomial x⁴-4x²-4x-1 given that two of its zeroes are 1+√2 and 1-√2. Since polynomials have complex roots in conjugate pairs, and the coefficients of the polynomial are real numbers, the other two roots must also be complex conjugates of each other.
The given roots suggest the factors (x - (1+√2)) and (x - (1-√2)) are part of the polynomial. Multiplying these two factors gives a quadratic factor of the polynomial. Let's find that quadratic factor.
First, expand the product of the two factors:
(x - (1+√2))(x - (1-√2)) = (x - 1 - √2)(x - 1 + √2) = x² - (1+√2)x - (1-√2)x + (1+√2)(1-√2) = x² - 2x + 1 - 2.
Now, simplifying gives us the quadratic factor:
x² - 2x - 1, which is part of the polynomial x⁴-4x²-4x-1.
To find the remaining two zeroes, we need to perform polynomial division of x⁴-4x²-4x-1 by x² - 2x - 1, or alternatively use a root-finding algorithm or numerical method, since the direct calculation might not yield a simplistic expression for the zeroes. However, the concept of solving polynomial equations remains the same.