If one of the zeroes of cubic polynomial x3 + ax2 + bx + c is - 1 then the product of the other two zeroes is -----
To find the product of the other two zeroes of the cubic polynomial, we can use Vieta's formulas. Vieta's formulas relate the coefficients of a polynomial to the roots of the polynomial.
In this case, let's assume the three zeroes of the cubic polynomial are -1, r, and s.
According to Vieta's formulas, the sum of the zeroes of a cubic polynomial is given by:
Sum of zeroes = (-1) + r + s = -a
Since one of the roots is -1, we have:
-1 + r + s = -a
Similarly, the product of the zeroes is given by:
Product of zeroes = (-1)(r)(s) = -c
Since we are looking for the product of the other two zeroes, we can express the product of the zeroes as:
(-1)(r)(s) = -c
Now, substituting the given values, we have:
(-1)(-1)(r)(s) = -c
(r)(s) = -c
Hence, the product of the other two zeroes of the cubic polynomial is -c.