Find p(0), p(1) and p(2) for of the fol | vedclass

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Question

$p(y)=y^{2}-y+1$

$\because $ $p(y)=y^{2}-y+1=(y)^{2}-y+1$

$	herefore$ $p(0)=(0)^{2}-(0)+1=0-0+1=1$

$p(1)=(1)^{2}-(1)+1=1-1+1=1$

$p(2)=(2)

Vedclass · Accepted Answer

Vedclass · Answer

$p(y)=y^{2}-y+1$

$\because $ $p(y)=y^{2}-y+1=(y)^{2}-y+1$

$	herefore$ $p(0)=(0)^{2}-(0)+1=0-0+1=1$

$p(1)=(1)^{2}-(1)+1=1-1+1=1$

$p(2)=(2)^{2}-2+1=4-2+1=3$

Find $p(0)$, $p(1)$ and $p(2)$ for of the following polynomials : $p(y)=y^{2}-y+1$

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