Question

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

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)p\left(y\right)=4+3y-y^2+5y^3 \\ \Rightarrow p\left(0\right)=\left(4+3\times 0-0^2+5\times 0^3\right) \end{gathered}$$
$$\begin{gathered} =\left(4+0-0+0\right) \\ =4 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)p\left(y\right)=4+3y-y^2+5y^3 \\ \Rightarrow p\left(2\right)=\left(4+3\times 2-2^2+5\times 2^3\right) \end{gathered}$$
$$\begin{gathered} =\left(4+6-4+40\right) \\ =46 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)p\left(y\right)=4+3y-y^2+5y^3 \\ \Rightarrow p\left(-1\right)=\left[4+3\times \left(-1\right)-{\left(-1\right)}^2+5\times {\left(-1\right)}^3\right] \end{gathered}$$
$$\begin{gathered} =\left(4-3-1-5\right) \\ =-5 \end{gathered}$$