If p(y) = 4 + 3y − y2 + 5y3,
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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