- Data Structure
- Networking
- RDBMS
- Operating System
- Java
- MS Excel
- iOS
- HTML
- CSS
- Android
- Python
- C Programming
- C++
- C#
- MongoDB
- MySQL
- Javascript
- PHP
- Physics
- Chemistry
- Biology
- Mathematics
- English
- Economics
- Psychology
- Social Studies
- Fashion Studies
- Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Verify whether the following are zeroes of the polynomial, indicated against them.
(i) $ p(x)=3 x+1, x=-\frac{1}{3} $
(ii) $ p(x)=5 x-\pi, x=\frac{4}{5} $
(iii) $ p(x)=x^{2}-1, x=1,-1 $
(iv) $ p(x)=(x+1)(x-2), x=-1,2 $
(v) $ p(x)=x^{2}, x=0 $
(vi) $ p(x)=l x+m, x=-\frac{m}{l} $
(vii) $ p(x)=3 x^{2}-1, x=-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}} $
(viii) $ p(x)=2 x+1, x=\frac{1}{2} $
To do:
We have to verify whether the given values are zeroes of the given polynomials.
Solution:
The zero of the polynomial is defined as any real value of $x$, for which the value of the polynomial becomes zero.
Therefore,
(i) \( p(x)=3 x+1, x=-\frac{1}{3} \)
$p(-\frac{1}{3})=3(-\frac{1}{3})+1$
$=-1+1$
$=0$
Therefore, $x=-\frac{1}{3}$ is the zero of the polynomial $p(x)=3 x+1$.
(ii) \( p(x)=5 x-\pi, x=\frac{4}{5} \)
$p(\frac{4}{5})=5(\frac{4}{5})-\pi$
$=4-\pi$
$≠0$
Therefore, $x=\frac{4}{5}$ is not the zero of the polynomial $p(x)=5x-\pi$.
(iii) \( p(x)=x^{2}-1, x=1,-1 \)
$p(1)=(1)^2-1$
$=1-1$
$=0$
$p(-1)=(-1)^2-1$
$=1-1$
$=0$
Therefore, $x=1,-1$ are the zeroes of the polynomial $p(x)=x^2-1$.
(iv) \( p(x)=(x+1)(x-2), x=-1,2 \)
$p(-1)=(-1+1)(-1-2)$
$=(0)(-3)$
$=0$
$p(2)=(2+1)(2-2)$
$=(3)(0)$
$=0$
Therefore, $x=-1,2$ are the zeroes of the polynomial $p(x)=(x+1)(x-2)$.
(v) \( p(x)=x^{2}, x=0 \)
$p(0)=(0)^2$
$=0$
Therefore, $x=0$ is the zero of the polynomial $p(x)=x^2$.
(vi) \( p(x)=l x+m, x=-\frac{m}{l} \)
$p(-\frac{m}{l})=l(-\frac{m}{l})+m$
$=-m+m$
$=0$
Therefore, $x=-\frac{m}{l}$ is the zero of the polynomial $p(x)=l x+m$.
(vii) \( p(x)=3 x^{2}-1, x=-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}} \)
$p(-\frac{1}{\sqrt3})=3(-\frac{1}{\sqrt3})^2-1$
$=3(\frac{1}{3})-1$
$=1-1$
$=0$
$p(-\frac{2}{\sqrt3})=3(-\frac{2}{\sqrt3})^2-1$
$=3(\frac{4}{3})-1$
$=4-1$
$=3$
$≠0$
Therefore, $x=-\frac{1}{\sqrt3}$ is the zero of the polynomial $p(x)=3x^2-1$ and $x=-\frac{2}{\sqrt3}$ is not the zero of the polynomial $p(x)=3x^2-1$.
(viii) \( p(x)=2 x+1, x=\frac{1}{2} \)
$p(\frac{1}{2})=2(\frac{1}{2})+1$
$=1+1$
$=2$
$≠0$
Therefore, $x=\frac{1}{2}$ is not the zero of the polynomial $p(x)=2 x+1$.