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) $p(x)=(x-1)(x+1)$

$p(0)=(0-1)(0+1)=(-1)(1)=-1$

$p(1)=(1-1)(1+1)=0(2)=0$

$p(2)=(2-1)(2+1)=1(3)=3$

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now