Question

Find a cubic polynomial whose zeroes are $\frac{1}{2}$, 1 and −3.

Answer 1

If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as
$x^3-\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x-abc$ ...(1)
Let $a=\frac{1}{2},b=1\mathrm{and}c=-3$
Substituting the values in (1), we get
$$\begin{gathered} x^3-\left(\frac{1}{2}+1-3\right)x^2+\left(\frac{1}{2}-3-\frac{3}{2}\right)x-\left(\frac{-3}{2}\right) \\ \Rightarrow x^3-\left(\frac{-3}{2}\right)x^2-4x+\frac{3}{2} \\ \Rightarrow 2x^3+3x^2-8x+3 \end{gathered}$$