Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(x+1\right)^{2}$.
$$\left(x-3\right)\left(x^{2}+2x+1\right)=0$$
Use the distributive property to multiply $x-3$ by $x^{2}+2x+1$ and combine like terms.
$$x^{3}-x^{2}-5x-3=0$$
By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $-3$ and $q$ divides the leading coefficient $1$. List all candidates $\frac{p}{q}$.
$$±3,±1$$
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 $x^{3}-x^{2}-5x-3$ by $x+1$ to get $x^{2}-2x-3$. Solve the equation where the result equals to $0$.
$$x^{2}-2x-3=0$$
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 $1$ for $a$, $-2$ for $b$, and $-3$ for $c$ in the quadratic formula.
