Answer

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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 function and equate it to zero to find the roots of the equation. The roots itself are the zeros of the polynomial.

**Complete step-by-step solution:**We need to find the zeros of the above quadratic equation given above $p\left( x \right)={{x}^{2}}+9x+18$.

Let’s understand by finding the zeros of the function.

Zeros of the polynomial can be defined as the points where the polynomial becomes zero.

Here, we have a quadratic function.

To find the zeros of the function we need to factorize the function.

Splitting the middle term, we get,

$\begin{align}

& p\left( x \right)={{x}^{2}}+9x+18 \\

& ={{x}^{2}}+6x+3x+18

\end{align}$

Taking the value of x common from the first two terms and 3 from the next terms we get,

$p\left( x \right)=x\left( x+6 \right)+3\left( x+6 \right)$

Taking $\left( x+6 \right)$ common we get,

$p\left( x \right)=\left( x+6 \right)\left( x+3 \right)$

We need to equate to zero, to find the points where the polynomial is zero.

Therefore, $\left( x+6 \right)\left( x+3 \right)=0$

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 $x + 3 = 0$,

$x = -3$.

The second point is $x + 6 = 0$,

$x = -6$,

**Hence, the two zeros of the given quadratic function are -3 and -6.**

**Note:**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 $-3-6=-9=\left( \dfrac{-9}{1} \right)$. Also, 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 $-3\times -6=18=\left( \dfrac{18}{1} \right)$. Hence, our answer is correct.

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