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

Solution :

Let the cubic polynomial be \(ax^3 + bx^2 + cx + d\), and its zeroes be \(\alpha\), \(\beta\) and \(\gamma\).

Then,  \(\alpha\) + \(\beta\) + \(\gamma\) = 2 = \(-(-2)\over 1\) = \(-b\over a\)

\(\alpha\)\(\beta\) + \(\beta\)\(\gamma\) + \(\gamma\)\(\alpha\) = – 7 = \(-7\over 1\) = \(c\over a\)

and \(\alpha\)\(\beta\)\(\gamma\) = -14 = \(-14\over 1\) = \(-d\over a\)

If a = 1, then b = -2, c = -7 and d = 14.

So, one cubic polynomial which satisfy the given conditions will be \(x^3 – 7x + 14\).