Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, 7, 14 respectively.

Let the cubic polynomial be denoted by p(x) = ax³ + bx² + cx + d, where a, b, c, and d are constants and a ≠ 0. Let α, β, and γ be the zeroes of the polynomial. We are given that:

Sum of zeroes (α + β + γ) = 2
Sum of the product of zeroes taken two at a time (αβ + βγ + γα) = 7
Product of zeroes (αβγ) = 14

For a cubic polynomial ax³ + bx² + cx + d = 0 with roots α, β, and γ, the following relationships hold:

α + β + γ = -b/a
αβ + βγ + γα = c/a
αβγ = -d/a

Comparing these relationships with the given information, we have:

-b/a = 2
c/a = 7
-d/a = 14

Let's assume a = 1 for simplicity. Then we get:

-b = 2 => b = -2
c = 7
-d = 14 => d = -14

Therefore, the cubic polynomial is:

p(x) = x³ - 2x² + 7x - 14

We can verify this. Let's use synthetic division or other methods to find the roots. One root is easily found by observation: if x=2, then

2³ - 2(2)² + 7(2) - 14 = 8 - 8 + 14 - 14 = 0

So x=2 is a root. Performing polynomial division (x³ - 2x² + 7x - 14) / (x-2), we get x² + 7. The roots of x² + 7 are ±√(-7) which are imaginary. Thus, the cubic polynomial with the given conditions is x³ - 2x² + 7x - 14.