Question

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Answer 1

answer-

Let the polynomial be ax3 + bx2 + cx + d and the zeroes be α, β and γ

Let the polynomial be ax3 + bx2 + cx + d and the zeroes be α, β and γThen, α + β + γ = -(-2)/1 = 2 = -b/a αβ + βγ + γα = -7 = -7/1 = c/a αβγ = -14 = -14/1 = -d/a

Let the polynomial be ax3 + bx2 + cx + d and the zeroes be α, β and γThen, 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

Let the polynomial be ax3 + bx2 + cx + d and the zeroes be α, β and γThen, 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 = 14So, one cubic polynomial which satisfy the given conditions will be x3 – 2×2 – 7x + 14

Answer 2

Answer:

x³-2x²-7x+14 is polynomial irrespective of the variable