Introduction:
To find all zeroes of the polynomial x^4-2root 2x^3 -18x^2 24 root 2x 72, we are given that 2root3 and -2root3 are two of its zeroes.

Using Factor Theorem:
We can use the factor theorem to find the other two zeroes. According to the factor theorem, if a polynomial P(x) has a factor (x-a), then P(a)=0.

Substituting the given zeroes:
Substituting x=2root3 in the polynomial, we get:
(2root3)^4-2root2(2root3)^3-18(2root3)^2+24root2(2root3)+72=0
Simplifying, we get:
64*12-16*24root3-648+48root6+72=0
Simplifying further, we get:
768-384root3-576+48root6+72=0
Simplifying more, we get:
-336+48root6-384root3=0
Rearranging, we get:
48root6-384root3=336
Dividing by 48, we get:
root6-8root3/3=7

Using Conjugate Root Theorem:
We know that the coefficients of the polynomial are all real. Therefore, if a+ib is a complex root of the polynomial, then its conjugate a-ib is also a root of the polynomial. We can use this fact to find the other two roots of the polynomial.

Finding the other two roots:
Since 2root3 is a root of the polynomial, its conjugate -2root3 is also a root of the polynomial. Therefore, we can write the polynomial as:
(x-2root3)(x+2root3)(ax^2+bx+c)
Expanding the polynomial, we get:
x^4-2root2x^3-18x^2+24root2x+72=ax^2(x^2-8)+bx(x^2-8)+c(x^2-8)
Simplifying, we get:
x^4-2root2x^3-18x^2+24root2x+72=(a+b+c)x^2-8(a+b+c)x+8(ab+ac+b^2+bc+c^2)
Since the polynomial has real coefficients, its complex roots must occur in conjugate pairs. Therefore, the other two roots of the polynomial must be complex conjugates of each other. Let us assume that these roots are p+qi and p-qi.

Using Vieta's Formulas:
Using Vieta's formulas, we know that:
p+qi+p-qi=2p=-b/a
(p+qi)(p-qi)=p^2+q^2=-c/a
Substituting these values in the equation above, we get:
x^4-2root2x^3-18x^2+24root2x+72=a(x^2+2px+p^2+q^2-8x)+8(bq+cq)
Simplifying, we get:
x^4-2root2x^3-18

Plz find the ans