Question

For each of the following polynomial, find $p\left(1\right),p\left(0\right)$ and $p\left(-2\right)$.

(i) $p\left(x\right)=x^3$ (ii) $p\left(y\right)=y^2-2y+5$ (iii) $p\left(x\right)=x^4-2x^2-x$

Answer 1

(i)
$$\begin{gathered} p\left(x\right)=x^3 \\ \therefore p\left(1\right)={\left(1\right)}^3=1 \\ p\left(0\right)={\left(0\right)}^3=0 \\ p\left(-2\right)={\left(-2\right)}^3=-8 \end{gathered}$$
(ii)
$$\begin{gathered} p\left(y\right)=y^2-2y+5 \\ \therefore p\left(1\right)={\left(1\right)}^2-2\times 1+5=1-2+5=4 \\ p\left(0\right)={\left(0\right)}^2-2\times 0+5=0-0+5=5 \\ p\left(-2\right)={\left(-2\right)}^2-2\times \left(-2\right)+5=4+4+5=13 \end{gathered}$$
(iii)
$$\begin{gathered} p\left(x\right)=x^4-2x^2-x \\ \therefore p\left(1\right)={\left(1\right)}^4-2\times {\left(1\right)}^2-1=1-2-1=-2 \\ p\left(0\right)={\left(0\right)}^4-2\times {\left(0\right)}^2-0=0-0-0=0 \\ p\left(-2\right)={\left(-2\right)}^4-2\times {\left(-2\right)}^2-\left(-2\right)=16-2\times 4+2=16-8+2=10 \end{gathered}$$