Verify whether the following are true or false.

(i) False, because zero of $x-3$ is $3 .$                  $[\because x-3=0 \Rightarrow x=3]$

(ii) True, because zero of $3 x+1$ is $-\frac{1}{3}$.  $\left[\because 3 x+1=0 \Rightarrow x=\frac{-1}{3}\right]$

(iii) False, because zero of $4-5 y$ is $\frac{4}{5}$.  $\left[\because 4-5 y=0 \Rightarrow y=\frac{4}{5}\right]$

(iv) True, now $t^{2}-2 t=t(t-2)$               

Hence, zeroes of $t^{2}-2 t$ are 0 and 2 .

(v) True, 

Now,

$y^{2}+y-6=y^{2}+3 y-2 y-6$                   [by splitting middle term]

$=y(y+3)-2(y+3)$

$=(y-2)(y+3)$

Hence, the zeroes of $y^{2}+y-6$ are 2 and $-3$.