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  • If one of the zeroes of the cubic polynomial x^3 + ax^2 + bx + c is –1, then the product of the other two zeroes is

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

(A) b – a + 1

(B) b – a – 1

(C) a – b + 1

(D) a – b –1

Answers (1)

Answer. [A]

Solution.  Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 3.

Here the given cubic polynomial is x^3 + ax^2 + bx + c

Let \alpha ,\beta ,\gamma are the zeroes of polynomial

\alpha =-1  (given)               …..(1)

put x = –1 in p(x)=x^3 + ax^2 + bx + c

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

0 = -1 + a - b + c ( Q=-1 is zero)

c = 1 - a + b

We know that

\alpha \beta \gamma =\frac{-d}{a}

Here, a = 1, b = a, c = b, d = c

So, \alpha \beta \gamma =\frac{-c}{1}

\beta \gamma =\frac{-(1-a+b)}{\alpha }

\beta \gamma =\frac{-(1-a+b)}{-1 } = 1 -a + b            ( using equation (1))

Hence the product of other two is b – a + 1.

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