iii) Let f(t)= 5t2+12t+7
= 5t2+7t+5t+7
= t(5t+7)+1(5t+7)
= (5t+7)(t+1)
So, the value of 5t2+12t+7 is zero when 5t + 7 = 0 or t + 1 = 0
i.e., when t= −75 or t=−1
So, the zeroes of 5t2+12t+7 are −75 and −1
∴ Sum of zeroes =−75−1=−125
= (−1)(Constant termCoefficient of t2)
and product of zeroes (−75)(−1)=75
= (−1)2(Constant termCoefficient of t2)
Hence, verified the relations between the zeroes and the coefficients of the polynomial