If one of the zeros of the cubic polynomial ax³ + bx2 + cx + d is 0, then the product of the other two zeros is

(a) $\frac{- c}{a}$
(b) $\frac{c}{a}$
(c) 0
(d) $\frac{- b}{a}$

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Solution

$(b) \frac{c}{a} Let α, β and 0 be the zeroes of a x^{3} + b x^{2} + c x + d . Then, sum of the products of zeroes taking two at at a time is given by (αβ + β \times 0 + α \times 0) = \frac{c}{a} = > αβ = \frac{c}{a} ∴ T he product of the other two zeroes is \frac{c}{a} .$

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