Solve x³-x²-4x+4=0: Finding All Solutions of a Cubic Equation

In the given equation:

$x^3-x^2-4x+4=0$

The simplest and fastest way to find the number of solutions,

will be simply to solve it, noting that the expression on the left side contains four different terms,

Additionally, note that we cannot factor out a common factor for all terms in the expression, since there is a free term,

therefore we will turn to factoring by groups:

We will consider two groups of terms in the given expression so that each group has two terms, we will choose the groups so that in each group only one term has an extreme power (in this case the third power and zero power - of the free term, are the extreme powers in the expression and therefore we will include in each group only one of these terms):

$x^3-x^2-4x+4=0 \\ \downarrow\\ \textcolor{red}{x^3-x^2}\textcolor{blue}{-4x+4}=0$

We'll continue and note that each group marked separately, can be factored by taking out a common factor, first we'll factor out a common factor from the first group, marked in red, and continue to the second group, marked in blue, we'll also factor it by taking out a common factor so that the expression inside the parentheses will be identical to the expression in parentheses in the second group (marked in red) in the following calculation the identical expression in parentheses will be emphasized with an underline:

$\textcolor{red}{x^3-x^2}\textcolor{blue}{-4x+4}=0 \\ \downarrow\\ \textcolor{red}{x^2\underline{(x-1)}}\textcolor{blue}{-4x+4}=0 \\ \downarrow\\ \textcolor{red}{x^2\underline{(x-1)}}\textcolor{blue}{-4\underline{(x-1)}}=0 \\$We'll continue and note that now- the complete expression on the left side can be factored further by taking out a common factor that is a binomial, meaning- we'll factor out the identical expression in parentheses that was marked with an underline, as a common factor, we'll do this:

$x^2\underline{(x-1)}-4\underline{(x-1)}=0 \\ \downarrow\\ \underline{(x-1)}(x^2-4)=0$We'll continue and note that the right expression in the product of expressions obtained on the left side, can also be factored,

this- using the perfect square binomial formula:

$a^2-b^2=(a+b)(a-b)$

Let's return to the equation we got and do this:

$(x-1)(x^2-4)=0 \\ (x-1)\textcolor{orange}{(x^2-2^2)}=0 \\ \downarrow\\ (x-1)\textcolor{orange}{(x+2)(x-2)}=0 \\$We got on the left side a product of expressions that must equal zero, remember that a product of expressions equals zero if and only if at least one of the expressions equals 0, therefore we get three simpler equations and solve them:

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

or:

$x+2=0\\ \downarrow\\ \boxed{x=-2}$

or:

$x-2=0\\ \downarrow\\ \boxed{x=2}$

Let's summarize then the equation solving steps:

$x^3-x^2-4x+4=0 \\ \downarrow\\ \textcolor{red}{x^3-x^2}\textcolor{blue}{-4x+4}=0 \\ \downarrow\\ \textcolor{red}{x^2\underline{(x-1)}}\textcolor{blue}{-4\underline{(x-1)}}=0 \\ \downarrow\\ \underline{(x-1)}\textcolor{orange}{(x^2-4)}=0 \\ (x-1)\textcolor{orange}{(x+2)(x-2)}=0 \\ \downarrow\\ x-1=0\rightarrow\boxed{x=1}\\ x+2=0\rightarrow\boxed{x=-2}\\ x-2=0\rightarrow\boxed{x=2}\\ \downarrow\\ \boxed{x=1,-2,2}$

Meaning- the given equation has three solutions,

Therefore the correct answer is answer C.