Question

Find a cubic polynomial with the sum, 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

Let 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 – 2×2 – 7x + 14

Answer 2

We know that the general form of a cubic polynomial is ax3 + bx2 + cx + d and the zeroes are α, β, and γ.

Let's look at the relation between sum, and product of its zeroes and coefficients of the polynomial.

α + β + γ = - b / aαβ + βγ + γα = c / aα x β x γ = - d / a

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

We know that,

α + β + γ = 2/1 = - b / a

αβ + βγ + γα = - 7/1 = c / a

α.β.γ = - 14/1 = - d / a

Thus, by comparing the coefficients we get, a = 1, then b = - 2, c = - 7 and d = 14

Now, substitute the values of a, b, c, and d in the cubic polynomial ax3 + bx2 + cx + d.

Hence the polynomial is x3 - 2x2 - 7x + 14.