Find $ p(0), p(1) $ and $ p(2) $ for each of the following polynomials:
(i) $ p(y)=y^{2}-y+1 $
(ii) $ p(t)=2+t+2 t^{2}-t^{3} $
(iii) $ p(x)=x^{3} $
(iv) $ p(x)=(x-1)(x+1) $
To do:
We have to find \( p(0), p(1) \) and \( p(2) \) for each of the given polynomials.
Solution:
To find the value of the polynomial $f(x)$ at $x=a$, we have to substitute $x=a$ in $f(x)$.
Therefore,
(i) \( p(y)=y^{2}-y+1 \)
$p(0) = (0)^{2}-(0)+1$
$= 0-0+1$
$= 1$
$p(1) = (1)^{2}-(1)+1$
$= 1-1+1$
$= 1$
$p(2) = (2)^{2}-(2)+1$
$= 4-2+1$
$= 3$
Hence, $p(0), p(1), p(2)$ for the given polynomial are $1,1$ and $3$ respectively.
(ii) \( p(t)=2+t+2 t^{2}-t^{3} \)
$p(0) = 2+0+2 (0)^{2}-(0)^{3}$
$= 2+2(0)-0$
$= 2+0$
$=2$
$p(1) = 2+1+2 (1)^{2}-(1)^{3}$
$= 3+2(1)-1$
$= 4$
$p(2) = 2+2+2(2)^2-(2)^3$
$= 4+2(4)-8$
$=-4+8$
$=4$
Hence, $p(0), p(1), p(2)$ for the given polynomial are $2,2$ and $4$ respectively.
(iii) \( p(x)=x^{3} \)
$p(0) = (0)^{3}$
$=0$
$p(1) =(1)^{3}$
$= 1$
$p(2) = (2)^{3}$
$= 8$
Hence, $p(0), p(1), p(2)$ for the given polynomial are $0,1$ and $8$ respectively.
(iv) \( p(x)=(x-1)(x+1) \)
$p(0) = (0-1)(0+1)$
$=(-1)(1)$
$=-1$
$p(1) =(1-1)(1+1)$
$=(0)(2)$
$=0$
$p(2) = (2-1)(2+1)$
$=(1)(3)$
$=3$
Hence, $p(0), p(1), p(2)$ for the given polynomial are $-1,0$ and $3$ respectively.
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