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# If the zeros of the quadratic polynomial ${{x}^{2}}+\left( a+1 \right)x+b$ are 2 and - 3, thenA.a = -7, b = -1B.a = 5, b = -1C.a = 2, b = -6D.a = 0, b = -6  Hint: For solving this problem, we obtain the sum and product of roots individually from the given equation and then place the values of variables in the generalized form to obtain the desired values of a and b.

In algebra, a quadratic function is a polynomial function with one or more variables in which the highest-degree term is of the second degree. A single-variable quadratic function can be stated as:
$f(x)=a{{x}^{2}}+bx+c,\quad a\ne 0$
If we have two zeros of a quadratic equation then the polynomial could be formed by using the simplified result which could be stated as:
${{x}^{2}}-(\alpha +\beta )x+\alpha \beta$, where $\alpha \text{ and }\beta$ are two zeroes of the equation.
According to the problem statement, we are given a quadratic polynomial ${{x}^{2}}+\left( a+1 \right)x+b$ having 2 and -3 as zeroes. Therefore, $\alpha =2\text{ and }\beta =-3$.
By using the above expression, the sum of zeroes could be expressed as
\begin{align} & -\left( \alpha +\beta \right)=a+1 \\ & -\left( 2+-3 \right)=a+1 \\ & -\left( -1 \right)=a+1 \\ & a=1-1 \\ & a=0 \\ \end{align}
The product of zeroes could be expressed
\begin{align} & \alpha \beta =b \\ & 2\times -3=b \\ & b=-6 \\ \end{align}
So, the respective values are a = 0 and b = -6.
Therefore, option (d) is correct.

Note: This problem could be alternatively solved by using the concept that zeros can also be expressed as factors. Now, we have -3 and 2 as zeros. Therefore, the polynomial would be represented as $\left( x+3 \right)\cdot \left( x-2 \right)={{x}^{2}}+x-6$. Now, on comparing this equation with the given problem, we obtain a = 0 and b = -6.

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