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If one of the zeroes of the cubic polynomial $x^{3} + a x^{2} + bx + c$ is $- 1$ then the product of the other two zeroes is

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1 - a + b

b - a - 1

a - b + 1

a - b - 1

answer is A.

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Detailed Solution

As it is given that if one of the zeroes of the cubic polynomial

x^{3} + a x^{2} + bx + c

- 1

.
Apply product of zeroes formula.
Consider the given polynomial to be

p (x) .

So,

p (x) = x^{3} + a x^{2} + bx + c

.
Given one of the zeroes of the polynomial is

- 1

x = - 1

Now, p(-1) = 0.

p (- 1) = {(- 1)}^{3} + a {(- 1)}^{2} + b (- 1) + c

\Rightarrow 0 = - 1 + a - b + c

\Rightarrow c = 1 - a + b

Using the product of zeroes formula,

Product of all zeroes = - constant coefficient of x 3

− 1 3 β γ = − c 1

β γ = c

So, the product of the other two zeroes is

1 - a + b

.
Hence, the correct option is 1.

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