Question

the value of p(0) for p(y)=y^2-y+1_______
a)0
b)1
c)2
d)4​

Answer 1

$p(y) = y^2 - y + 1$

Now,

$p(0)$ will be equal to p(y) where y = 0, i.e.,

$p(0) = 0^2 - 0 + 1= 1$

Therefore, the value of $p(0)$ is (b) 1.

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