Question

Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y)=y2−y+1
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Answer 1

$\huge\star{\underline{\mathtt{\red{A}\pink{N}\green{S}\blue{W}\purple{E}\orange{R}}}}⋆$

$p(y) = y^2–y+1$

$∴p(0) = (0)^2−(0)+1=1$

$p(1) = (1)^2–(1)+1=1$

$p(2) = (2)^2–(2)+1=3$

Answer 2

$\huge\underline\red{SOLUTION}$

p(0)

♠p(y)=y²−(y+1)

♠p(0)=0²−(0+1)

♠p(0)=0−1

♠p(0)= -1

p(1)

♠p(y)=y²−(y+1)

♠p(1)=1²−(1+1)

♠p(1)=1−2

♠p(1)= −1

p(2)

♠p(y)=y²−(y+1)

♠p(2)=2²−(2+1)

♠p(2)=4−3

♠p(2)=1

Hope it helps...