Question

find the zeroes of the polynomial t^2+8t+16 and verify the relationship between the zeroes and cofficients
it's urgent please​

Answer 1

Answer:

the zeroes are (-4) and (-4)

Explanation:

Given,

t^2+8t+16

so the factors of 16 are 1,2,4,8,16

t^2+4t+4t+16 (factors of 16 whose sum is 8)

t(t+4)+4(t+4)

(t+4) (t+4). t=( -4),t= (-4)

so relation between zeroes and coefficients

a+b= -b/a. = ( -4)+(-4) =( -8). = -b/a. =. 8

a*b= c/a. =.( -4)*(-4) =. 16. = c/a = 16

Hence ,proved