Solve the Cubic Equation: x³-x²-4x+4=0

To solve this problem, we need to factor the cubic polynomial equation $x^3 - x^2 - 4x + 4 = 0$. We'll begin by applying the Rational Root Theorem, which suggests that possible rational roots are factors of the constant term (4) divided by factors of the leading coefficient (1). This gives us potential roots: $\pm 1, \pm 2, \pm 4$.

Let's test these possible roots by substituting them into the polynomial:

• For $x = 1$, the polynomial evaluates to $1^3 - 1^2 - 4 \times 1 + 4 = 1 - 1 - 4 + 4 = 0$. Thus, $x = 1$ is a root.
• For $x = -1$, it evaluates to $(-1)^3 - (-1)^2 - 4(-1) + 4 = -1 - 1 + 4 + 4 = 6$. Thus, $x = -1$ is not a root.
• For $x = 2$, it evaluates to $2^3 - 2^2 - 4 \times 2 + 4 = 8 - 4 - 8 + 4 = 0$. Thus, $x = 2$ is a root.
• For $x = -2$, it evaluates to $(-2)^3 - (-2)^2 - 4(-2) + 4 = -8 - 4 + 8 + 4 = 0$. Thus, $x = -2$ is a root.

From these calculations, we identified $x = 1$, $x = 2$, and $x = -2$ as roots of the polynomial.

The polynomial can be factored as $(x - 1)(x - 2)(x + 2) = 0$. Solving each factor for zero, we obtain the roots $x = 1$, $x = 2$, and $x = -2$.

Therefore, the correct answer from the given choices is Answers a and c, which correspond to the roots $x = \pm 2$ and $x = 1$.