Question

4u^+8u product and sum of zeros​

Answer 1

Answer:

sum of zeroes = - coefficient of u / coefficient of u^2

= -(8)/4

=-2

product of zeroes = constant term / coefficient of u^2

= 0/4

= 0

Answer 2

$\large{\colorbox{black}{\underline{\underline{\text{\red{☆answer☆}}}}}}$

4u^2 + 8u

4u ( u + 2 )

4u + 0 = 0 , u + 2 = 0

u = 0 , u = -2

HENCE ZEROES ARE : 0 , -2

let α = 0 , β = -2

Sum of zeroes

= α + β

= 0 + ( -2 )

= -2

Product of zeroes

= αβ

= 0 × ( -2 )

= 0

$\large{\colorbox{pink}{\underline{\underline{\text{\red{☆like \: \: the \: \: answer☆}}}}}}$

$\large{\colorbox{black}{\underline{\underline{\text{ \red{☆brainlist \: \: pls☆}}}}}}$