Question

Find a cubic polynomial whose zeroes are 3, – 1, – 1​

Answer 1

$\underline{\bf{Given}}$

• Zeroes of Cubic polynomial:- 3,-1,-1

$\underline{\bf{To Find}}$

• Cubic Polynomial

$\\Let , \\ \alpha = 3 \: \: \: \: \: \beta = - 1 \: \: \: \: \: \gamma = - 1$

$\\ \text {{if a β γ are the zeros of a cubic polynomial f (x) \: then}}$

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

$\sf{f(x) = {x}^{3} - (3 + ( - 1) + ( - 1)) {x}^{2} + ( 3 \times - (1) + ( -1 ) \times ( - 1)+ ( - 1) \times 3)x - (3 \times ( - 1)( - 1) }$

$\sf{f(x) = {x}^{3} - (3 - 2) {x}^{2} + ( - 3 + 1 - 3)x - 3}$

$\sf{f(x) = {x}^{3} - {x}^{2} - 5x - 3}$

$\text{\sf{The cubic polynomial whose zeroes are 3, – 1, – 1}}\\\bf{\pink{f(x) = x³-x²-5x-3}}$