Question

What are the possible rational roots of $x^3-5x^2-4x+20=0$ and then determine the rational roots?

Answer 1

Finding rational roots of $x^3-5x^2-4x+20=0$:

The possible rational roots are the factors of the constant $20$ divided by the factors of the leading coefficient $1$.

Possible rational roots:$\pm 1,\pm 2,\pm 4,\pm 5,\pm 10,\pm 20$

Factoring the polynomial:

$$\begin{gathered} \Rightarrow x^3-5x^2-4x+20=0 \\ \Rightarrow x^2\left(x-5\right)-4\left(x-5\right)=0 \\ \Rightarrow \left(x^2-4\right)\left(x-5\right)=0 \\ \Rightarrow \left(x-2\right)\left(x+2\right)\left(x-5\right)=0 \end{gathered}$$

This gives $x=2or-2or5$

Hence, the rational roots are $-2,2,5$.