Question

Find the cubic polynomial with the sum of zeroes, sum of product of its zeroes taken 2 at a time and product of its zeroes as 0, -7 and -6 respectively.
A. \[{x^3} - 7x - 6\]
B. \[{x^3} - 7{x^2} + 6\]
C. \[{x^3} - 7x + 6\]
D. \[{x^3} - 7{x^2} - 6\]

Answer 1

Hint: If we consider a cubic polynomial \[a{x^3} + b{x^2} + cx + d = 0\] , which is also the general form of a cubic polynomial and also we are assuming that \[\alpha ,\beta ,\gamma \] are the roots of a cubic polynomial then \[\alpha + \beta + \gamma = - \dfrac{b}{a},\alpha \beta + \beta \gamma + \alpha \gamma = \dfrac{c}{a},\alpha \beta \gamma = - \dfrac{d}{a}\] We will use this to gain relation between b, c, and d to get the required polynomial.

Complete step-by-step answer:
We are given that
\[\begin{array}{l}
\alpha + \beta + \gamma = - \dfrac{b}{a} = 0\\
\alpha \beta + \beta \gamma + \alpha \gamma = \dfrac{c}{a} = - 7\\
\alpha \beta \gamma = - \dfrac{d}{a} = - 6
\end{array}\]
Where a, b, c are the coefficients of \[{x^3},{x^2},x\] respectively and d is the constant part.
So we will substitute the values and get the final equations as
\[\begin{array}{l}
 \Rightarrow - \dfrac{b}{a} = 0..............................................(i)\\
 \Rightarrow \dfrac{c}{a} = - 7.................................................(ii)\\
 \Rightarrow - \dfrac{d}{a} = - 6..................................................(iii)
\end{array}\]
On solving equation (i) we will get
\[\begin{array}{l}
 \Rightarrow - \dfrac{b}{a} = 0\\
 \Rightarrow - b = 0\\
 \Rightarrow b = 0
\end{array}\]
From equation (ii) we will get
\[\begin{array}{l}
 \Rightarrow \dfrac{c}{a} = - 7\\
 \Rightarrow c = - 7a
\end{array}\]
From equation (iii) we will get
\[\begin{array}{l}
 \Rightarrow - \dfrac{d}{a} = - 6\\
 \Rightarrow \dfrac{d}{a} = 6\\
 \Rightarrow d = 6a
\end{array}\]
Now we know that the general cubic polynomial is of the form \[a{x^3} + b{x^2} + cx + d = 0\]
So let us put the values of b, c and d as we get it from the above equations.
\[\begin{array}{l}
\therefore a{x^3} + b{x^2} + cx + d = 0\\
 \Rightarrow a{x^3} + 0 \times {x^2} + ( - 7a)x + 6a = 0\\
 \Rightarrow a{x^3} - 7ax + 6a = 0\\
 \Rightarrow a\left( {{x^3} - 7x + 6} \right) = 0\\
 \Rightarrow {x^3} - 7x + 6 = 0
\end{array}\]

So, the correct answer is “Option C”.

Note: Forming the correct equations are the key step and to observe where to put the values as in this case all values containing a is put because it was clear that a could be taken in common and then we can cancel it out as on the other side of the polynomial there is a zero only.