Question

If k is any non-zero constant, then the cubic polynomial whose sum of zeroes, sum of product of its zeroes taken two at a time and product of its zeroes are 3, -7 and -3 respectively is?

Answer 1

Hint: We write the general form of a quadratic equation and assume three zeros of the equation. Write three equations using the given information and use the formula of forming a quadratic equation using the zeroes of the quadratic polynomial.
* If \[\alpha ,\beta ,\gamma \] are three zeros of a cubic polynomial and we are given the value of \[\alpha + \beta + \gamma = S\]; \[\alpha \beta + \beta \gamma + \gamma \alpha = T\] and \[\alpha \beta \gamma = P\], then the cubic polynomial can be formed as \[k({x^3} - S{x^2} + Tx - P)\]

Complete step-by-step solution:
Let us take general form of a cubic polynomial i.e.\[a{x^3} + b{x^2} + cx + d = 0\]
Let the three zeros of the polynomial be \[\alpha ,\beta ,\gamma \].
We know that if
\[\alpha + \beta + \gamma = S\]................… (1)
\[\alpha \beta + \beta \gamma + \gamma \alpha = T\]...............… (2)
\[\alpha \beta \gamma = P\].....................… (3)
 Then the cubic polynomial can be formed as \[k({x^3} - S{x^2} + Tx - P)\]
We are given sum of zeroes of cubic polynomial is 3
\[ \Rightarrow \alpha + \beta + \gamma = 3\]
Substitute the value of \[\alpha + \beta + \gamma = S\] from equation (1)
\[ \Rightarrow S = 3\]................… (4)
We are given sum of product of its zeroes taken two at a time is -7
\[ \Rightarrow \alpha \beta + \beta \gamma + \gamma \alpha = - 7\]
Substitute the value of \[\alpha \beta + \beta \gamma + \gamma \alpha = T\] from equation (2)
\[ \Rightarrow T = - 7\].................… (5)
We are given product of zeroes of cubic polynomial is -3
\[ \Rightarrow \alpha \beta \gamma = - 3\]
Substitute the value of \[\alpha \beta \gamma = P\] from equation (3)
\[ \Rightarrow P = - 3\]...............… (6)
Then substitute the value of S, T and P from equations (4), (5) and (6) respectively in \[k({x^3} - S{x^2} + Tx - P)\]
Then the cubical polynomial is \[k({x^3} - 3{x^2} + ( - 7)x - ( - 3))\]
We know products of one negative number and one positive number are negative numbers and the product of two negative numbers is a positive number.
Then the cubical polynomial is \[k({x^3} - 3{x^2} - 7x + 3)\]

\[\therefore \]Cubical polynomial formed is \[k({x^3} - 3{x^2} - 7x + 3)\]

Note: Students make mistake by solving this question by trying to solve three equations formed using the zeroes of the cubic polynomial, but this is a very long and complex process.