Solve the Cubic Equation: x(x²-7x-8)=0 Using Factoring

Let's solve the given equation:

$x(x^2-7x-8)=0$

Let's try to factor the expression inside of the parentheses:

$x^2-7x-8$

Note that in this expression the coefficient of the squared term is 1, therefore, we can (try to) factor the expression on the left side using quick trinomial factoring:

Let's look for a pair of numbers whose product equals the free term in the expression, and whose sum equals the coefficient of the first-degree term, meaning two numbers $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=-8\\ m+n=-7\\$From the first requirement mentioned, that is - from the multiplication, we notice that the product of the numbers we're looking for needs to be negative, therefore we can conclude that the two numbers have different signs, according to multiplication rules, and now we'll remember that the possible factors of 8 are 2 and 4, or 8 and 1, fulfilling the second requirement mentioned, along with the fact that the numbers we're looking for have different signs will lead to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=-8\\ n=1 \end{cases}$

Therefore we'll factor the expression on the left side of the equation:

$x^2-7x-8\\ \downarrow\\ (x-8)(x+1)$

Now let's return to the original equation and apply the factoring that we completed earlier:

$x(x^2-7x-8)=0 \\ \downarrow\\ x(x-8)(x+1)=0$

Remember that the product of expressions will yield 0 only if at least one of the multiplied expressions equals zero,

Therefore we obtain three simple equations and solve them by isolating the variable on one side:

$\boxed{x=0}$

or:

$x-8=0\\ \boxed{x=8}$

or:

$x+1=0\\ \boxed{x=-1}$

Let's summarize the solution of the equation:

$x(x^2-7x-8)=0 \\ \downarrow\\ x(x-8)(x-3)=0 \\ \downarrow\\ \boxed{x=0}\\ x-8=0\rightarrow\boxed{x=8}\\ x+1=0\rightarrow\boxed{x=-1}\\ \downarrow\\ \boxed{x=0,8,-1}$

Therefore the correct answer is answer C.