Question

If one of the zeroes of the cubic polynomial x 3 + px² + qx + r is -1, then the
product of the other two zeroes is
(a) p + q + 1
(b) p-q- 1
(c) q – p + 1
(d) q – p – 1

Answer 1

Answer:

(c) q- p + 1

Step-by-step explanation:

Let p(x) = x³ + px² + qx + r

p(-1) = 0

(-1)³+p(-1)²+q(-1) + r = 0

-1+ p -q + r = 0

r = q- p + 1

$let \: \: zeroes \: \: be \: \: \alpha , \beta \: \: and \: \: \gamma . \\ \alpha \beta \gamma = \frac{ - d}{a} \\ \alpha \beta ( - 1) = \frac{ - r}{1} \\ - \alpha \beta = - r \\ \alpha \beta = r \\ \alpha \beta = q - p + 1$

Answer 2

Step-by-step explanation:

r=q-p+1..

it is ur answer