Find p(0), p(1) and p(2) for each of the following polynomials: <br/> <br/> (i) $p(y)=y^{2}-y+1$<br/> <br/> (ii) $p(t)=2+t+2 t^{2}-t^{3}$<br/> <br/>(iii) $p(x)=x^{3}$ <br/> <br/>(iv) $p(x)=(x-1)(x+1)$
Solution:
(i) $p(y)=y^{2}-y+1$
$p(0)=(0)^{2}-(0)+1=1$
$p(1)=(1)^{2}-(1)+1=1$
$p(2)=(2)^{2}-(2)+1=3$
(ii) $p(t)=2+t+2 t^{2}-t^{3}$
$p(0)=2+0+2(0)^{2}-(0)^{3}=2$
$p(1)=2+(1)+2(1)^{2}-(1)^{3}$
$=2+1+2-1=4$
$p(2)=2+2+2(2)^{2}-(2)^{3}$
$=2+2+8-8=4$
(iii) $p(x)=x^{3}$
$p(0)=(0)^{3}=0$
$p(1)=(1)^{3}=1$
$p(2)=(2)^{3}=8$
(iv) $p(x)=(x-1)(x+1)$
$p(0)=(0-1)(0+1)=(-1)(1)=-1$
$p(1)=(1-1)(1+1)=0(2)=0$
$p(2)=(2-1)(2+1)=1(3)=3$
(i) $p(y)=y^{2}-y+1$
$p(0)=(0)^{2}-(0)+1=1$
$p(1)=(1)^{2}-(1)+1=1$
$p(2)=(2)^{2}-(2)+1=3$
(ii) $p(t)=2+t+2 t^{2}-t^{3}$
$p(0)=2+0+2(0)^{2}-(0)^{3}=2$
$p(1)=2+(1)+2(1)^{2}-(1)^{3}$
$=2+1+2-1=4$
$p(2)=2+2+2(2)^{2}-(2)^{3}$
$=2+2+8-8=4$
(iii) $p(x)=x^{3}$
$p(0)=(0)^{3}=0$
$p(1)=(1)^{3}=1$
$p(2)=(2)^{3}=8$
(iv) $p(x)=(x-1)(x+1)$
$p(0)=(0-1)(0+1)=(-1)(1)=-1$
$p(1)=(1-1)(1+1)=0(2)=0$
$p(2)=(2-1)(2+1)=1(3)=3$
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