Question

(19) Find a cubic polynomial with the sum of zeros, sum of the product of its zeros taken two at a time
and the product of its zeros as 2,-7,-14 respectively.​

Answer 1

Answer:-

$\mathsf{x^3 -2x^2 -7x -14 }$

Given :-

Sum of zeroes = 2

Sum of the product of its zeroes taken two at a time = -7

Product of its zeroes = -14

To find :-

The required cubic polynomial.

Solution:-

Let $\alpha , \beta$$\gamma$ be the zeroes of required polynomial.

A/Q

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

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

$\alpha \beta \gamma = - 14$

The required cubic polynomial is given by :-

$\text{ x^3 - (sum of zeroes) x^2 + (sum of products of zeroes)x - (product of zeroes)}$

Put the given value,

${x}^{3} - ( \alpha + \beta + \gamma ) {x}^{2} + ( \alpha \beta + \beta \gamma + \gamma \alpha )x - ( \alpha \beta \gamma )$

$\mathsf{x^3 - (2) x^2 + (-7) x-14}$

$\mathsf{ x^3 -2x^2 -7x -14}$

hence,

The required cubic polynomial will be $\mathsf{x^3 -2x^2 -7x -14 }$

Answer 2

$\mathfrak{\huge{\green{\underline{\underline{AnswEr :}}}}}$

Given :

» Sum of Zeroes ${(\alpha + \beta + \gamma )}$ = 2

» Sum of Product ot its Zeroes taken at two times ${( \alpha\beta + \beta\gamma + \gamma\alpha )}$ = -7

» Product of Zeroes ${(\alpha\beta\gamma)}$ = -14

Solution :

Let the Zeroes of Cubic Polynomial be $\bold { \alpha , \beta \: and \: \gamma }$

⚘ ${x}^{3} - (sum \: of \: zeroes) {x}^{2} + (sum \: of \: product \: of \: zeroes \: taken \: at \: two \: times)x \: - (product \: of \: zeroes)$

⚘ ${x}^{3} - ( \alpha + \beta + \gamma ) {x}^{2} + ( \alpha \beta + \beta \gamma + \gamma \alpha )x - ( \alpha \beta \gamma )$

⚘ ${x}^{3} - (2) {x}^{2} + ( - 7)x - 14$

⚘ ${x}^{3} - 2 {x}^{2} - 7x - 14$

❖ Hence the Cubic Polynomial will be $\mathscr\blue{{x}^{3} - 2 {x}^{2} - 7x - 14}</p><p>$