The zero of the quadratic polynomial $x^{2} + (a + 1) x + b$ are $2$ and $- 3$ , then
a) $a = - 7, b = - 1$
b) $a = 5, b = - 1$
c) $a = 2, b = - 6$
c) $a = 0, b = - 6$

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Solution

Let $p (x) = x^{2} + (a + 1) x + b$
Given that, $2$ and $- 3$ are the zeroes of the quadratic polynomial $p (x)$
$∴ p (2) = 0 and p (- 3) = 0,$

Solving for $p (2)$ ,
$\Rightarrow 2^{2} + (a + 1) (2) + b = 0$
$\Rightarrow 4 + 2 a + 2 + b = 0$
$\Rightarrow 2 a + b = - 6 \dots (i)$

Solving for $p (- 3),$
$\Rightarrow (- 3)^{2} + (a + 1) (- 3) + b = 0$
$\Rightarrow 9 - 3 a - 3 + b = 0$
$\Rightarrow 3 a - b = 6 \dots (i i)$

On adding Eqs. $(i) a n d (i i)$ , we get
$5 a = 0 \Rightarrow a = 0$
Substituting the value of $a$ in Eq. $(i)$ , we get
$x \times 0 + b = - 6 \Rightarrow b = - 6$
So, the required values are $a = 0$ and $b = - 6$ .

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