Question:
If $p(y)=4+3 y-y^{2}+5 y^{3}$, find
(i) $p(0)$,
(ii) $p(2)$,
(iii) $p(-1)$
Solution:
(i) $p(y)=4+3 y-y^{2}+5 y^{3}$
$\Rightarrow p(0)=\left(4+3 \times 0-0^{2}+5 \times 0^{3}\right)$
$=(4+0-0+0)$
$=4$
(ii) $p(y)=4+3 y-y^{2}+5 y^{3}$
$\Rightarrow p(2)=\left(4+3 \times 2-2^{2}+5 \times 2^{3}\right)$
$=(4+6-4+40)$
$=46$
(iii) $p(y)=4+3 y-y^{2}+5 y^{3}$
$\Rightarrow p(-1)=\left[4+3 \times(-1)-(-1)^{2}+5 \times(-1)^{3}\right]$
$=(4-3-1-5)$
$=-5$
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