Question

Find a cubic 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

Let the polynomial be $a{x}^3+b{x}^2+cx+d$ and the zeroes be $\alpha ,\beta ,\gamma$.
Sum of the polynomial = $\alpha +\beta +\gamma =\frac{2}{1}=\frac{-b}{a}$

$\Rightarrow \frac{-b}{a}=\frac{2}{1}$ ---(1)
Sum of the product of its zeroes taken two at a time = $\alpha \beta +\beta \gamma +\alpha \gamma =\frac{7}{1}=\frac{c}{a}$

$\Rightarrow \frac{c}{a}=\frac{7}{1}$ ----(2)
Product of the root = $\alpha \beta \gamma =\frac{14}{1}=\frac{-d}{a}$

$\Rightarrow \frac{-d}{a}=\frac{14}{1}$ ----(3)

From (1), (2) and (3), $a=1$ ---(i)

[Since, in each case denominator in LHS is $a$ and denominator in RHS is $1$]

Now, from (1),

$\frac{-b}{a}=\frac{2}{1}$

$\Rightarrow \frac{-b}{1}=\frac{2}{1}$

$\Rightarrow b=-2$ --(ii)

from (2),

$\frac{c}{a}=\frac{7}{1}$

$\Rightarrow \frac{c}{1}=\frac{7}{1}$

$\Rightarrow c=7$ ---(iii)

from (2),

$\frac{-d}{a}=\frac{14}{1}$

$\Rightarrow \frac{-d}{1}=\frac{14}{1}$

$\Rightarrow d=-14$ ---(iv)
i.e., $a=1,b=-2,c=7,d=-14.$

Hence, the required polynomial

$=a{x}^3+b{x}^2+cx+d$

$=x^3-2x^2+7x-14$

(OR)

If $\alpha ,\beta ,\gamma$ are the zeros of a cubic polynomial then the polynomial is,

$P\left(x\right)=x^3-(\alpha +\beta +\gamma )x^2+(\alpha \beta +\beta \gamma +\alpha \gamma )x-\alpha \beta \gamma$

It is given that,

$\alpha +\beta +\gamma =2$

$\alpha \beta +\beta \gamma +\alpha \gamma =7$

$\alpha \beta \gamma =14$

$\Rightarrow P\left(x\right)=x^3-2x^2+7x-14$