If the sum of zeroes of p(x)=(a+1) $x^{2}$ +(2a+3)x+(3a+4) is -1.Find its product.

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Solution

$P (x) = (a + 1) x^{2} + (2 a + 3) x + (3 a + 4)$
Sum of root (s) = -1
$(\frac{- (c o f f . o f x)}{c o f f . o f x^{2}})$
$\Rightarrow \frac{- (2 a + 3)}{a + 1} = - 1$
$- (2 a + 3) = - (a + 1)$
$- 2 = a$
Product of roots $= \frac{3 a + 4}{a + 1} = \frac{3 (- 2) + 4}{- 2 + 1} = 2$
$(\frac{C o n s t a n t}{c o f f o f n^{2}})$

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