Question

Find a cubic polynomial with the sum, sum of the product of its zeros taken two at a time, and the product of its zeros as 3, -1, -3 respectively.

Answer 1

Hint: To solve this problem, one should have theoretical knowledge of cubic equations and the relationship between their coefficients and roots. The formulas that will be used to verify this relationship are-\[\alpha +\beta +\gamma =-\dfrac{b}{a}\], \[\alpha \beta +\gamma \beta +\gamma \alpha =\dfrac{c}{a}\] and \[\alpha \beta \gamma =-\dfrac{d}{a}\]. So, using the information given in the question and these three formulas, we will first reform the cubic equation and then we will obtain the equation of the cubic polynomial.

Complete step by step solution:
Let the equation be $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$
We will divide both sides by a in the above equation, and then apply the formula for the relationship between the roots and the coefficients, which is given by
\[\alpha +\beta +\gamma =-\dfrac{b}{a}\], \[\alpha \beta +\gamma \beta +\gamma \alpha =\dfrac{c}{a}\] and \[\alpha \beta \gamma =-\dfrac{d}{a}\].
As it is given that the sum, sum of the product of its zeros taken two at a time, and the product of its zeros as 3, -1, -3 which means
\[\alpha +\beta +\gamma =3\], \[\alpha \beta +\gamma \beta +\gamma \alpha =-1\] and \[\alpha \beta \gamma = -3\].
So, here, we can see that we can substitute these formulas in the equation-
${{x}^{3}}+\dfrac{b}{a}{{x}^{2}}+\dfrac{c}{a}x+\dfrac{d}{a}=0$
\[{{x}^{3}}-\left( \alpha +\beta +\gamma \right){{x}^{2}}+\left( \alpha \beta +\gamma \beta +\gamma \alpha \right)x-\alpha \beta \gamma \text{=0 }\]
We put the given values to get
\[{{x}^{3}}-\left( 3 \right){{x}^{2}}+\left( -1 \right)x+\left( 3 \right)\text{=0 }\]
On simplifying, we get
\[{{x}^{3}}-3{{x}^{2}}-x+3=0\]
Thus, the required cubic polynomial with sum, sum of the product of its zeros taken two at a time, and the product of its zeros 3, -1 and -3 respectively is equals to \[{{x}^{3}}-3{{x}^{2}}-x+3=0\].

Note: Always remember that general equation of cubic polynomial is $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$, where a, b, c and d belongs to set of real numbers and also keep the logic remember that number of roots of polynomial is equals to highest power of x in polynomial. The most common mistake here is that students often forget the negative sign in the sum and the product of roots formula. Try not to make any calculation errors while solving the question.