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# Form a quadratic polynomial whose zeroes are 3 and -1.  Verified
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Hint:Here first we will assume the given zeros to be $\alpha$ and $\beta$ then we will find the sum and product of the zeroes and substitute the respective values to get the desired quadratic equation.

Formula used:For a standard quadratic equation $a{x^2} + bx + c = 0$ and if $\alpha$ and $\beta$ are the roots/zeros of the equation then
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:-
${x^2} - 2x - 3 = 0$
${x^2} - 2x - 3 = 0$
${x^2} - 2x - 3$.
$p\left( x \right) = {x^2} - \left( {{\text{sum of zeroes}}} \right)x + \left( {{\text{product of zeroes}}} \right)$.