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The Factor Theorem is a fundamental theorem in algebra that connects roots of a polynomial to its factors. It states that a polynomial \( f(t) \) has a factor \( t-a \) if and only if \( f(a) = 0 \). This means that if, after substituting \( a \) into the polynomial, the result is zero, \( t-a \) is indeed a factor. To apply this theorem, first identify the value of \( a \) from \( t-a \), which will be our test value for the polynomial. In our problem, we replace \( t \) with 3, because the divisor is \( t-3 \). Substituting 3 into the polynomial, we compute \( 3^5 - 3 \, (3^4) - 3^2 - 6 \). If the result is zero, \( t-3 \) is a factor; if not, it isn't. This aligns directly with the essence of the Factor Theorem.

Synthetic division is a shortcut method of polynomial division, especially useful when dividing by linear factors. It simplifies the process by focusing on coefficients. Here's how it works:

• Write the coefficients of the polynomial in descending order, filling in zeros for missing terms.
• Place the root (from \( t-a \)) to the left of the coefficients. In this case, it's 3 because of \( t-3 \).
• Bring down the leading coefficient to the bottom row.
• Multiply this coefficient by the root and add it to the next coefficient.
• Repeat the process for each coefficient, continually multiplying by the root and adding, until all coefficients are processed.

The last number you write is the remainder. If it is zero, then the divisor is a factor. For our polynomial, after conducting synthetic division, the remainder was \(-15\), which is not zero, confirming \( t-3 \) is not a factor.

Finding the roots of a polynomial is equivalent to solving when the polynomial equals zero. These roots are significant because each root represents a solution to the polynomial equation and can also indicate factors according to the Factor Theorem.When using synthetic division or evaluating a polynomial for possible roots, focus on substituting values derived from potential linear factors. In the problem at hand, if \( t=3 \) were a root, then \( t-3 \) would be a factor, as the polynomial would equate to zero for \( t=3 \).However, for this polynomial, \( 3 \) is not a root, indicated by the non-zero remainder when using synthetic division: \(-15\). Therefore, there's no zero at \( t=3 \), and \( t-3 \) is not a root. Understanding this intertwined relationship between roots and polynomial factors is crucial for solving polynomial equations effectively.