Question

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

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

Answer 1

$$\begin{gathered} \left(\text{a}\right)\frac{-b}{a} \\ \text{Let}\mathrm{\alpha }\text{, 0 and 0 be the zeroes of}a{x}^3+b{x}^2+cx+d=0. \\ \text{Then sum of the zeroes=}\frac{-b}{a} \\ =>\alpha +0+0=\frac{-b}{a} \\ =>\alpha =\frac{-b}{a} \\ \text{Hence, the third zero is}\frac{-b}{a}\text{.} \end{gathered}$$