How do you find a polynomial of degree $3$ that has zeros of $- 3$ , $0$ , $1$ ?

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Solution

How to find a polynomial of degree $3$ that has zeros $- 3$ , $0$ , $1$ :
Let, $α = - 3, β = 0 a n d γ = 1$
The formula of the cubic polynomial when three roots $(α, β, and γ)$ are given is
$k (x^{3} - (α + β + γ) x^{2} + (α β + β γ + γ α) x - α β γ)$ , where $k \neq 0$
Substitute the values in the formula:
$k (x^{3} - (- 3 + 0 + 1) x^{2} + ((- 3) (0) + (0) (1) + (1) (- 3)) x - (- 3) (0) (1)) = k (x^{3} - (- 2) x^{2} + (0 + 0 - 3) x - 0) = k (x^{3} + 2 x^{2} - 3 x)$
Hence, the polynomial of degree $3$ is $\begin{array}{rcl} k (x^{3} + 2 x^{2} - 3 x) . \end{array}$

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