Question

Find p(0) and p(1) for each of the following polynomials..​

Answer 1

Answer:

$i) \: \: \: \: \: p(0) = 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: p(1) = 1 \\ \\ \\ \\ \\ \\ ii) \: \: \: \: \: \: p(0) = 2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: p(1) = 6$

Step-by-step explanation:

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

$ii) \: \: \: \: p(t) = 2 + t + 2 {t}^{2} + {t}^{3} \\ \\ p(0) = 2 + 0 + 0 + 0 \\ = 2 \\ \\ \\ \\ p(1) = 2 + 1 + 2 + 1 \\ \\ = 6$