If p(y) = 4 + 3y − y² + 5y³, find (i) p(0), (ii) p(2), (iii) p(−1).

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Solution

$(i) p (y) = 4 + 3 y - y^{2} + 5 y^{3} \Rightarrow p (0) = (4 + 3 \times 0 - 0^{2} + 5 \times 0^{3})$
$= (4 + 0 - 0 + 0) = 4$

$(ii) p (y) = 4 + 3 y - y^{2} + 5 y^{3} \Rightarrow p (2) = (4 + 3 \times 2 - 2^{2} + 5 \times 2^{3})$
$= (4 + 6 - 4 + 40) = 46$

$(iii) p (y) = 4 + 3 y - y^{2} + 5 y^{3} \Rightarrow p (- 1) = [4 + 3 \times (- 1) - {(- 1)}^{2} + 5 \times {(- 1)}^{3}]$
$= (4 - 3 - 1 - 5) = - 5$

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