Question

Find a cubical 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.

Answer 1

Hint: In this question, we need to write the polynomial equation in terms of their coefficients. Then from the given conditions we can get the values of the coefficients. Now, substitute these coefficient values in the polynomial equation to get the result.

Complete step by step answer:
POLYNOMIAL: An expression of the form \[{{a}_{0}}{{x}^{n}}+{{a}_{1}}{{x}^{n-1}}+{{a}_{2}}{{x}^{n-2}}+......+{{a}_{n-1}}x+{{a}_{n}},\] where \[{{a}_{0}},{{a}_{1}},{{a}_{2}}.......,{{a}_{n-1}},{{a}_{n}}\] are real numbers and n is a non-negative integer, is called a polynomial in the variable x. Polynomials in variable x are generally denoted by \[f\left( x \right)\].
CUBIC POLYNOMIAL: A cubic polynomial is a polynomial having degree three.
The general form of a cubic polynomial is given by \[a{{x}^{3}}+b{{x}^{2}}+cx+d=0\]
ZERO OF A POLYNOMIAL: A real number \[\alpha \] is a zero of the polynomial p(x), if and only if \[p\left( \alpha \right)=0\]
Now, let us assume the zeroes of the cubic polynomial as
\[\alpha ,\beta ,\gamma \]
Now, from the cubic polynomial \[a{{x}^{3}}+b{{x}^{2}}+cx+d=0\] we have,
Sum of the roots of the cubic polynomial is given by
\[\alpha +\beta +\gamma =\dfrac{-b}{a}\]
Now, sum of the product of the two roots at a time is given by
\[\alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{c}{a}\]
Now, product of all the roots is given by
\[\alpha \beta \gamma =\dfrac{-d}{a}\]
Now, from the given values in the question we have,
\[\alpha +\beta +\gamma =2\]
\[\alpha \beta +\beta \gamma +\gamma \alpha =-7\]
\[\alpha \beta \gamma =-14\]
Now, on comparing these values with the respective formulae we get,
\[\Rightarrow \dfrac{-b}{a}=2\]
\[\Rightarrow \dfrac{c}{a}=-7\]
\[\Rightarrow \dfrac{-d}{a}=-14\]
Now, let us write the coefficients b, c, d in terms of a
\[\Rightarrow -b=2a\]
Now, on multiplying with -1 on both sides we get,
\[\therefore b=-2a\]
\[\therefore c=-7a\]
\[\Rightarrow -d=-14a\]
Now, on multiplying with -1 on both sides we get,
\[\therefore d=14a\]
Now, let us substitute these values of coefficients in the cubic polynomial.
\[\Rightarrow a{{x}^{3}}+b{{x}^{2}}+cx+d=0\]
Now, on substituting the respective values of b, c, d in terms of a we get,
\[\Rightarrow a{{x}^{3}}+\left( -2a \right){{x}^{2}}+\left( -7a \right)x+14a=0\]
Now, this can also be written as
\[\Rightarrow a{{x}^{3}}-2a{{x}^{2}}-7ax+14a=0\]
Let us now divide with a on both sides to get the cubic polynomial
\[\therefore {{x}^{3}}-2{{x}^{2}}-7x+14=0\]
Note:
Instead of directly getting the values of coefficients in terms of other from the given conditions we can also do it by first calculating the zeroes of the polynomial by solving the three equations. Then from the values of the zeroes we can get the coefficients but it will be a lengthy process.
It is important to note that to get the equation of the cubic polynomial with the given conditions of the zeroes we need not calculate the values of the zeroes. We can directly get the relation between the coefficients and substitute in the assumed polynomial equation to get the result.