Question

Find the zero of the polynomial in each of the following cases; p(y)=3y+1

Answer 1

Answer:

to find zeroes,

p(y) = 0

=> 3y +1 = 0

=> 3y = -1

=> y = -⅓

so, the zero of p(x) is -⅓.

Answer 2

Answer:

$\huge\bold{\underline{\underline{Solution}}}$

$p(y) = 3y + 1 \\ \\ \\ \implies3y + 1 = 0 \\ \\ \\ \implies3y = - 1 \\ \\ \\ \implies \: y = \frac{ - 1}{3} \\ \\ \\ so \: the \: zeros \: of \: p(y) = \frac{ - 1}{3}$