91Ó°ÊÓ

The Factor Theorem is a useful tool in polynomial simplification. It helps us determine whether a particular expression is a factor of a polynomial. If substituting a certain value into the polynomial results in zero, that value corresponds to a factor. Here, we want to check if \(x-2\) is a factor of \(x^3 - 2x^2 - 4x + 8\). By plugging \(x = 2\) into the polynomial, we compute \(2^3 - 2(2)^2 - 4(2) + 8 = 0\). Because the result is zero, \(x-2\) is indeed a factor.

This step is crucial because it informs our next steps, like polynomial division or synthetic division, aiding in breaking down complex polynomials into simpler pieces.

Synthetic Division is a streamlined method to divide polynomials. It's especially useful when dealing with linear divisors such as \(x-c\). In this problem, we've confirmed \(x-2\) is a factor, so we use synthetic division to divide the polynomial \(x^3 - 2x^2 - 4x + 8\) by \(x-2\).

Here’s a quick overview of how it works:

• Write down the coefficients: \(1, -2, -4, 8\).
• Place \(2\) (from \(x-2\) set equal to zero) on the left side.
• Perform the division by bringing down the first coefficient, multiplying, adding, and repeating these steps across all coefficients.

As a result, we find the quotient: \(x^2 + 0x - 4\). A key advantage of synthetic division is its efficiency and speed compared to long division.

The Difference of Squares is a pattern that helps us factor certain quadratic expressions quickly. The formula is \(a^2 - b^2 = (a-b)(a+b)\). In our polynomial simplification, after using synthetic division, we are left with the expression \(x^2 - 4\).

This can be factored as:

• Identify that \(4\) is \(2^2\), making \(x^2 - 4\) match the format of a difference of squares.
• Apply: \(x^2 - 4 = (x-2)(x+2)\).

By recognizing this pattern, we can efficiently factor the quadratic, which helps us simplify the polynomial further.

A Removable Discontinuity occurs when a point on the graph is not defined due to a factor that cancels out in the simplification process. In this simplified function, \(f(x) = \frac{x^3 - 2x^2 - 4x + 8}{x-2}\), the factor \(x-2\) cancels out from both numerator and denominator.

Once canceled, our function becomes \(f(x) = x^2 - 4\), but it must be noted that originally \(x = 2\) was a point of discontinuity, not appearing in the simplified version. To complete understanding, we compute the limit at \(x=2\) using the simplified function. Calculating \(\lim_{{x \to 2}} f(x) = 2^2 - 4 = 0\), the limit shows the value the function approaches is \(0\).

In graphing, this concept is represented by a hole at \((2, 0)\), and it's crucial to mark these points as they impact the graph interpretation and understanding.