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
\(b-a+1\)
\(b-a-1\)
\(a-b+1\)
\(a-b-1\)
Step 1: Use the fact that \(-1\) is a root.
Substitute \(x=-1\) in the polynomial:
\((-1)^3 + a(-1)^2 + b(-1) + c = 0\)
\(-1 + a - b + c = 0\)
So, \(c = b - a + 1\).
Step 2: Recall the relationship of roots with coefficients.
For a cubic polynomial \(x^3 + ax^2 + bx + c\), if the roots are \(\alpha, \beta, \gamma\), then:
\(\alpha + \beta + \gamma = -a\)
\(\alpha\beta + \beta\gamma + \gamma\alpha = b\)
\(\alpha\beta\gamma = -c\)
Step 3: Express product of the other two zeroes.
Let the roots be \(-1, r, s\).
Then \((-1)rs = -c\)
So, \(rs = c\).
Step 4: Substitute the value of \(c\).
Since \(c = b - a + 1\), we get
\(rs = b - a + 1\).
Final Answer: Option (A) \(b - a + 1\).