By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $5$ and $q$ divides the leading coefficient $5$. List all candidates $\frac{p}{q}$.
$$±1,±5,±\frac{1}{5}$$
Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.
$$X=1$$
By Factor theorem, $X-k$ is a factor of the polynomial for each root $k$. Divide $5X^{4}-3X^{3}-4X^{2}-3X+5$ by $X-1$ to get $5X^{3}+2X^{2}-2X-5$. Solve the equation where the result equals to $0$.
$$5X^{3}+2X^{2}-2X-5=0$$
By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $-5$ and $q$ divides the leading coefficient $5$. List all candidates $\frac{p}{q}$.
$$±1,±5,±\frac{1}{5}$$
Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.
$$X=1$$
By Factor theorem, $X-k$ is a factor of the polynomial for each root $k$. Divide $5X^{3}+2X^{2}-2X-5$ by $X-1$ to get $5X^{2}+7X+5$. Solve the equation where the result equals to $0$.
$$5X^{2}+7X+5=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. Substitute $5$ for $a$, $7$ for $b$, and $5$ for $c$ in the quadratic formula.
