Question

help help help help help help help help help help help​

Answer 1

(iv) 4u2+8u

⇒ 4u(u+2)

Therefore, zeroes of polynomial equation 4u2 + 8u are (0, -2).

Sum of zeroes = 0+(-2) = -2 = -(8/4) = = -(Coefficient of u)/(Coefficient of u2)

Product of zeroes = 0×-2 = 0 = 0/4 = (Constant term)/(Coefficient of u2 )

(v) t2–15

⇒ t2 = 15 or t = ±√15

Therefore, zeroes of polynomial equation t2 –15 are (√15, -√15)

Sum of zeroes =√15+(-√15) = 0= -(0/1)= -(Coefficient of t) / (Coefficient of t2)

Product of zeroes = √15×(-√15) = -15 = -15/1 = (Constant term) / (Coefficient of t2 )

Follow me

Make me as brainlist please