If one of the zeros of the cubic polynomial x^{3} + ax^{2} + bx + c is –1 then the product of the other two zeros is

It is given in the question that,

- 1 is the zero of the given polynomial x^{3} + ax^{2} + bx + c

∴ (- 1)^{3} + a × (- 1)^{2} + b × (- 1) + c = 0

a – b + c + 1 = 0

c = 1 – a + b

Also, the product of all zeros is:

× (- 1) = - c

= c

= 1 – a + b

Hence, option C is correct

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