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The Rational Root Theorem is a powerful tool for finding the possible rational roots of a polynomial equation. This theorem states that if a polynomial with integer coefficients has a rational root \(\frac{p}{q}\), then \(p\) is a factor of the constant term and \(q\) is a factor of the leading coefficient.
In our problem, we have the polynomial \(P(x) = x^{5} - 3x^{4} + 12x^{3} - 28x^{2} + 27x - 9\). Here, the constant term is \(-9\) and the leading coefficient is \(1\).
This means potential rational roots are the factors of \(-9\) divided by the factors of \(1\).

• Factors of \(-9\) are: \(\pm 1, \pm 3, \pm 9\).
• Factors of \(1\) are: \(\pm 1\).

Thus, the potential rational roots are: \(\pm 1, \pm 3, \pm 9\).
We then test each of these possible roots by substituting them into the polynomial equation to see if they yield zero. Any value that returns zero is a valid root.

Synthetic division is a streamlined method for dividing a polynomial by a linear factor of the form \(x - c\). It is particularly useful for verifying potential roots obtained from the Rational Root Theorem since it simplifies the division process.
To use synthetic division, set up a grid with the coefficients of the polynomial and the potential root \(c\). For example, if we found that \(x=1\) is a root, we would use \(1\) in our synthetic division.

• List the coefficients of \(P(x)\): \(1, -3, 12, -28, 27, -9\).
• The potential root \(c = 1\).

We perform synthetic division which involves multiplying, adding, and repeating these steps for each coefficient. If the remainder is zero, \(x-c\) is a factor, confirming \(c\) as a root. The process also reduces the degree of the polynomial, allowing us to continue testing and dividing if needed, using remaining coefficients.
Performing synthetic division multiple times helps break down the original polynomial into simpler forms, ultimately aiding in identifying all roots.

Complex roots occur when solving a quadratic or higher-degree polynomial gives solutions involving the imaginary unit \(i\), which is described as \(i^2 = -1\). These roots often appear in conjugate pairs such as \(a+bi\) and \(a-bi\).
In solving the given problem, after several rounds of synthetic division using the root \(x=1\), we ended up with a quadratic equation \(x^2 + 9 = 0\).
To solve \(x^2 + 9 = 0\), we set \(x^2 = -9\), and take the square root of both sides:

• \(x = \pm \sqrt{-9}\)
• \(x = \pm 3i\)

These indicate the complex roots of the polynomial. Thus, alongside the repeated real root \(x = 1\), we discover the complex roots \(x = 3i\) and \(-3i\).
These complex roots provide a complete set of solutions when combined with the real roots, fulfilling the degree of the original polynomial equation.