Use the distributive property to multiply $x^{2}+4x-5$ by $x-5$ and combine like terms.
$$x^{3}-x^{2}-25x+25=-24\left(x-1\right)$$
Use the distributive property to multiply $-24$ by $x-1$.
$$x^{3}-x^{2}-25x+25=-24x+24$$
Add $24x$ to both sides.
$$x^{3}-x^{2}-25x+25+24x=24$$
Combine $-25x$ and $24x$ to get $-x$.
$$x^{3}-x^{2}-x+25=24$$
Subtract $24$ from both sides.
$$x^{3}-x^{2}-x+25-24=0$$
Subtract $24$ from $25$ to get $1$.
$$x^{3}-x^{2}-x+1=0$$
By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $1$ and $q$ divides the leading coefficient $1$. List all candidates $\frac{p}{q}$.
$$±1$$
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 $x^{3}-x^{2}-x+1$ by $x-1$ to get $x^{2}-1$. Solve the equation where the result equals to $0$.
$$x^{2}-1=0$$
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 $a$, $0$ for $b$, and $-1$ for $c$ in the quadratic formula.
