Question

Find the zeroes of the polynomial of x^2-2

Answer 1

Answer:

Zeros of the polynomial are √2 and -√2

Step-by-step explanation:

$\mathsf{(x)^{2} -(\sqrt{2}) ^{2}=0}$

$\mathsf{(x+\sqrt{2})(x-\sqrt{2})=0}$

$\mathsf{[a^{2}-b^{2}=(a+b)(a-b)]}$

$\mathsf{x=\sqrt{2} \ or\ x=-\sqrt{2}}$

Answer 2

Topic - Polynomials

Polynomials: An expression of the form

$\quad a_{0}x^{n}+a_{1}x^{n-1}+...+a_{n-1}x+a_{n}$,

where $a_{0},\:a_{1},\: ...,\:a_{n}$ are given numbers (real or complex), $n$ is a negative integer and $x$ is a variable, is called a polynomial in $x$ or a rational integral function of $x$.

Zero of a polynomial: $\alpha$ is said to be a zero of order $r$ of the polynomial $f(x)$ if $(x-\alpha)^{r}$ is a factor of $f(x)$ while $(x-\alpha)^{r+1}$ is not a factor of $f(x)$.

Solution: Let the given polynomial be represented as

$\quad\quad f(x)=x^{2}-2$

To find the zeroes of $f(x)$, we equate it with zero $(0)$. Then

$\quad f(x)=0$

$\Rightarrow x^{2}-2=0$

$\Rightarrow x^{2}-(\sqrt{2})^{2}=0$

$\Rightarrow (x+\sqrt{2})\:(x-\sqrt{2})=0$

$\therefore$ either $x+\sqrt{2}=0$

$\quad\quad$ or, $x-\sqrt{2}=0$

$\Rightarrow x=-\sqrt{2},\:\sqrt{2}$

Answer: Therefore the zeroes of the polynomial $(x^{2}-2)$ are $\pm \sqrt{2}$.