Product of Remaining Zeroes of a Cubic Polynomial

Video Explanation

Question

If one of the zeroes of the cubic polynomial

\[ f(x) = x^3 + ax^2 + bx + c \]

is \(-1\), find the product of the other two zeroes.

Options:

(a) \(b – a + 1\)
(b) \(b – a – 1\)
(c) \(a – b + 1\)
(d) \(a – b – 1\)

Solution

Step 1: Write Relations Between Zeroes and Coefficients

Let the zeroes of the polynomial be:

\[ -1,\; \alpha,\; \beta \]

For a cubic polynomial:

\[ \alpha + \beta + (-1) = -a \]

\[ \alpha\beta + \beta(-1) + (-1)\alpha = b \]

Step 2: Simplify the Relations

From the sum of zeroes:

\[ \alpha + \beta = -a + 1 \]

From the sum of products of zeroes taken two at a time:

\[ \alpha\beta – (\alpha + \beta) = b \]

Step 3: Substitute the Value of \(\alpha + \beta\)

\[ \alpha\beta – (-a + 1) = b \]

\[ \alpha\beta + a – 1 = b \]

\[ \alpha\beta = b – a + 1 \]

Conclusion

The product of the other two zeroes is:

\[ \boxed{b – a + 1} \]

Hence, the correct option is (a).