To factor the expression, solve the equation where it equals to $0$.
$$x^{4}-8x^{3}+5x^{2}+8x-6=0$$
By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $-6$ and $q$ divides the leading coefficient $1$. List all candidates $\frac{p}{q}$.
$$±6,±3,±2,±1$$
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 $x^{4}-8x^{3}+5x^{2}+8x-6$ by $x-1$ to get $x^{3}-7x^{2}-2x+6$. To factor the result, solve the equation where it equals to $0$.
$$x^{3}-7x^{2}-2x+6=0$$
By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $6$ and $q$ divides the leading coefficient $1$. List all candidates $\frac{p}{q}$.
$$±6,±3,±2,±1$$
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 $x^{3}-7x^{2}-2x+6$ by $x+1$ to get $x^{2}-8x+6$. To factor the result, solve the equation where it equals to $0$.
$$x^{2}-8x+6=0$$
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 $a$, $-8$ for $b$, and $6$ for $c$ in the quadratic formula.
