Find p(0) , p(1) and p(2) for of following polynomials : p(y)=y^{2}-y+1

English

Gujarati

Hindi

2. Polynomials

easy

Find $p(0)$, $p(1)$ and $p(2)$ for of the following polynomials : $p(y)=y^{2}-y+1$

Option A

Option B

Option C

Option D

Solution

$p(y)=y^{2}-y+1$

$\because $ $p(y)=y^{2}-y+1=(y)^{2}-y+1$

$\therefore$ $p(0)=(0)^{2}-(0)+1=0-0+1=1$

$p(1)=(1)^{2}-(1)+1=1-1+1=1$

$p(2)=(2)^{2}-2+1=4-2+1=3$

Standard 9

Mathematics

Share

Similar Questions

Without actually calculating the cubes, find the value of each of the following : $(-12)^{3}+(7)^{3}+(5)^{3}$

medium

Find the zero of the polynomial : $p(x) = 3x$

easy

Divide the polynomial $3 x^{4}-4 x^{3}-3 x-1$ by $x-1$.

hard

Verify whether the following are zeroes of the polynomial, indicated against them.

$p(x)=3 x^{2}-1,\,x=-\,\frac{1}{\sqrt{3}},\, \frac{2}{\sqrt{3}}$

easy

Write the following cubes in the expanded form : $(5 p-3 q)^{3}$

medium