Question

Let the cubic polynomial be ( a x^{3}+b x^{2}+c x+d ) Let ( alpha, beta, gamma ) be the zeroes of the required cubic polynomial,
then
( alpha+beta+gamma=frac{-b}{a} ; alpha beta+beta gamma+gamma alpha=frac{c}{a} ; ) and ( alpha beta gamma=frac{-d}{a} )
Now, according to the question, ( alpha+beta+gamma=2 )
( Rightarrow frac{-b}{a}=2 Rightarrow frac{b}{a}=-2 )
( alpha beta+beta gamma+gamma alpha=-7 Rightarrow frac{mathrm{c}}{mathrm{a}}=-7 )
And ( quad alpha beta gamma=-14 Rightarrow frac{-d}{a}=-14 ; frac{d}{a}=14 )
If ( a=1, ) then ( b=-2, c=-7, d=14 )
Substituting the values of ( a, b, c ) and ( d ) in (i), we get the required cubic polynomial ( x^{3}-2 x^{2}-7 x+14 )
This polynomial satisfies all the given conditions

# Form a cubic polynomial having sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes are 2, -7 and -14 respectively.

Solution