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

Answer:

${x}^{3} - 2 {x}^{2} - 7x +14 = 0$

Step-by-step explanation:

$\alpha + \beta + \gamma = 2 \: \: \: \: \: ...(i) \\ \\ \alpha \beta + \beta \gamma + \gamma \alpha = - 7 \: \: \: \:... (ii) \\ \\ \alpha \beta \gamma = - 14 \: \: \: \: \: ...(iii) \\ \\ {x}^{3} - ( \alpha + \beta + \gamma ) {x}^{2} + ( \alpha \beta + \beta \gamma + \gamma \alpha )x - \alpha \beta \gamma = 0 \\ \\ {x}^{3} - 2 {x}^{2} - 7x +14 = 0.$

Answer 2

Answer:

et the polynomial be ax3 + bx2 + cx + d and the zeroes be α, β and γ Then, α + β + γ = -(-2)/1 = 2 = -b/a αβ + βγ + γα = -7 = -7/1 = c/a αβγ = -14 = -14/1 = -d/a∴ a = 1, b = -2, c = -7 and d = 14 So, one cubic polynomial which satisfy the given conditions will be x3 - 2x2  - 7x + 14

Step-by-step explanation: