Given that one of the zeroes of the cubic polynomial \(ax^3+bx^2+cx+d\) is zero, the product of the other two zeroes is
\(-\dfrac{c}{a}\)
\(\dfrac{c}{a}\)
0
\(-\dfrac{b}{a}\)
Step 1: Write down the roots.
The polynomial is \(ax^3 + bx^2 + cx + d\).
It is given that one root is \(0\). Let the other two roots be \(\alpha\) and \(\beta\).
Step 2: Use relations between coefficients and roots.
For a cubic polynomial \(ax^3 + bx^2 + cx + d\) with roots \(r_1, r_2, r_3\):
\(r_1 + r_2 + r_3 = -\dfrac{b}{a}\)
\(r_1r_2 + r_2r_3 + r_3r_1 = \dfrac{c}{a}\)
\(r_1r_2r_3 = -\dfrac{d}{a}\)
Step 3: Substitute the known root.
Take \(r_1 = 0\), \(r_2 = \alpha\), \(r_3 = \beta\).
Then
\(r_1r_2 + r_2r_3 + r_3r_1 = 0 \cdot \alpha + \alpha\beta + 0 \cdot \beta\)
\(= \alpha\beta\).
Step 4: Compare with the formula.
We know \(r_1r_2 + r_2r_3 + r_3r_1 = \dfrac{c}{a}\).
So \(\alpha\beta = \dfrac{c}{a}\).
Answer: The product of the other two zeroes is \(\dfrac{c}{a}\).