Question

chose a quadratic polynomial whose some and the product of zeros is 3and 2 respectively A y²+2y-3 B y²+3y+2 C Y²-3Y +2 D y²-3y + 1​

Answer 1

Step-by-step explanation:

y² + 4√3y - 15 = 0

Where, cofficient of x² = 1 , cofficient of x = 4√3 , constant or cofficient of x⁰ = -15

now finding the zeroes of the polynomial by splitting middle term.

y² + 4√3y - 15 = 0

y² + 5√3y - √3y - 15 = 0

y(y + 5√3) - √3(y + 5√3) = 0

(y - √3)(y + 5√3) = 0

y = √3 , y = -5√3

Now,

sum of zeroes = (-cofficient of x) / (cofficient of x²)

√3 + (-5√3) = -4√3/1

-4√3 = -4√3

product of zeroes = cofficient of x⁰/cofficient of x²

(√3) * (-5√3) = (-15)/1

-15 = -15