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Find the zeros and the numbers of zeros of the polynomial ${{x}^{2}}-3$ and verify the relationship between the zeros and the coefficients.

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Hint: In order to solve this problem, we need to understand the concept of zeros of the polynomial. Zeros of the polynomial can be defined as the points where the polynomial becomes zero.
We need to factorize the quadratic equation and equate to zero to find the roots of the equation.
The roots itself are the zeros of the polynomial. In order to verify, we can use the property of the quadratic equation. In the equation $a{{x}^{2}}+bx+c=0$ , the sum of the roots is given by $\dfrac{-b}{a}$. In the equation $a{{x}^{2}}+bx+c=0$, the product of the roots is given by $\dfrac{c}{a}$.

Complete step-by-step solution:
We need to find the zeros of the above quadratic equation given above $p\left( x \right)={{x}^{2}}-3$.
Let’s understand by finding the zeros of the equation.
Zeros of the polynomial can be defined as the points where the polynomial becomes zero.
Here, we have a quadratic equation.
To find the zeros of the equation we need to factorize the equation.
As there is no middle term we can add and subtract the middle term with the coefficients of the square root of the constants.
Splitting the middle term, we get,
$\begin{align}
  & p\left( x \right)={{x}^{2}}-3 \\
 & ={{x}^{2}}+\sqrt{3}x-\sqrt{3}x-3
\end{align}$
Taking the value of x common from the first two terms and $\sqrt{3}$ from the next terms we get,
\[p\left( x \right)=x\left( x+\sqrt{3} \right)-\sqrt{3}\left( x+\sqrt{3} \right)\]
Taking $\left( x+\sqrt{3} \right)$ common we get,
$p\left( x \right)=\left( x+\sqrt{3} \right)\left( x-\sqrt{3} \right)$
We need to equate to zero, to find the points where the polynomial is zero.
Therefore, $\left( x+\sqrt{3} \right)\left( x-\sqrt{3} \right)=0$
This equation is satisfied only when anyone term is zero.
So, we have two points of x where the polynomial is zero.
The first point is $x+\sqrt{3}=0$ ,
$x=-\sqrt{3}$
The second point is $x-\sqrt{3}=0$ ,
$x=\sqrt{3}$
Hence, the two roots of the equation are $-\sqrt{3}$ and $\sqrt{3}$ .
 In order to check this, we can use the property of the quadratic equation. In the equation $a{{x}^{2}}+bx+c=0$ , the sum of the roots is given by $\dfrac{-b}{a}$ . In this question the sum of roots is $-\sqrt{3}+\sqrt{3}=0=0$.
In the equation $a{{x}^{2}}+bx+c=0$, the product of the roots is given by $\dfrac{c}{a}$.
In this question the product of roots is $-\sqrt{3}\times \sqrt{3}=3=3$.
Hence, our answer is correct.

Note: We can also factorize with a different approach. We can use the identity of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ .So, we in our question we can write as ${{x}^{2}}-3=\left( x-\sqrt{3} \right)\left( x+\sqrt{3} \right)$.
We need to remember that there is a negative sign while verifying with the coefficients of the quadratic equation, the sum of the roots is equal to $\dfrac{-b}{a}$.