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The sum of the zeroes is given by:-

\[\alpha + \beta = \dfrac{{ - b}}{a}\]

The product of the zeroes is given by:-

\[\alpha \beta = \dfrac{c}{a}\]

The given zeros are:-

3 and -1

Let \[\alpha = 3\] and \[\beta = - 1\]

Now since we know that for standard quadratic equation \[a{x^2} + bx + c = 0\] and if \[\alpha \] and \[\beta \] are the roots/zeroes of the equation then

The sum of the zeroes is given by:-

\[\alpha + \beta = \dfrac{{ - b}}{a}\]

Hence putting in the known values we get:-

\[3 - 1 = \dfrac{{ - b}}{a}\]

Solving it further we get:-

\[\dfrac{2}{1} = \dfrac{{ - b}}{a}\]

Comparing the values we get:-

\[

- b = 2 \\

\Rightarrow b = - 2 \\

\]

And, \[a = 1\]

Now we also know that the product of the zeroes is given by:-

\[\alpha \beta = \dfrac{c}{a}\]

Now putting in the known values we get:-

\[3\left( { - 1} \right) = \dfrac{c}{a}\]

Simplifying it further we get:-

\[\dfrac{{ - 3}}{1} = \dfrac{c}{a}\]

Comparing the values we get:-

\[c = - 3\] and \[a = 1\]

Putting the values of a, b and c in the standard quadratic equation we get:-

The standard quadratic equation is given by:

\[a{x^2} + bx + c = 0\]

Putting in the values of a, b, c we get:-

\[\left( 1 \right){x^2} + \left( { - 2} \right)x + \left( { - 3} \right) = 0\]

Simplifying it further we get:-

\[{x^2} - 2x - 3 = 0\]

Hence the quadratic equation is:

\[{x^2} - 2x - 3 = 0\]

The quadratic polynomial is:

\[{x^2} - 2x - 3\].

\[p\left( x \right) = {x^2} - \left( {{\text{sum of zeroes}}} \right)x + \left( {{\text{product of zeroes}}} \right)\].