The Rational Root Theorem states that any rational root of the polynomial \( f(x) = 3x^4 - 2x^3 + 74x^2 - 50x - 25 \) must be a factor of the constant term divided by a factor of the leading coefficient.
- Constant term: \(-25\)
- Leading coefficient: \(3\)
Possible rational zeros are:
\[
\pm 1, \pm 5, \pm 25, \pm \frac{1}{3}, \pm \frac{5}{3}, \pm \frac{25}{3}
\]
We will use synthetic division to test these possible rational zeros.
\[
\begin{array}{r|rrrrr}
1 & 3 & -2 & 74 & -50 & -25 \\
& & 3 & 1 & 75 & 25 \\
\hline
& 3 & 1 & 75 & 25 & 0 \\
\end{array}
\]
Since the remainder is \(0\), \(x = 1\) is a root.
Using synthetic division, we factor \( f(x) \) by \( (x - 1) \):
\[
f(x) = (x - 1)(3x^3 + x^2 + 75x + 25)
\]
Now, we need to find the rational zeros of \( 3x^3 + x^2 + 75x + 25 \).
Possible rational zeros are:
\[
\pm 1, \pm 5, \pm 25, \pm \frac{1}{3}, \pm \frac{5}{3}, \pm \frac{25}{3}
\]
\[
\begin{array}{r|rrrr}
-1 & 3 & 1 & 75 & 25 \\
& & -3 & 2 & -77 \\
\hline
& 3 & -2 & 77 & -52 \\
\end{array}
\]
The remainder is not \(0\), so \( x = -1 \) is not a root.
\[
\begin{array}{r|rrrr}
-5 & 3 & 1 & 75 & 25 \\
& & -15 & 70 & -725 \\
\hline
& 3 & -14 & 145 & -700 \\
\end{array}
\]
The remainder is not \(0\), so \( x = -5 \) is not a root.
\[
\begin{array}{r|rrrr}
-\frac{1}{3} & 3 & 1 & 75 & 25 \\
& & -1 & \frac{2}{3} & -25 \\
\hline
& 3 & 0 & 75 & 0 \\
\end{array}
\]
Since the remainder is \(0\), \( x = -\frac{1}{3} \) is a root.
Using synthetic division, we factor \( 3x^3 + x^2 + 75x + 25 \) by \( (3x + 1) \):
\[
3x^3 + x^2 + 75x + 25 = (3x + 1)(x^2 + 25)
\]
The quadratic polynomial \( x^2 + 25 \) can be factored as:
\[
x^2 + 25 = (x + 5i)(x - 5i)
\]
\[
\boxed{x = 1, x = -\frac{1}{3}}
\]
\[
\boxed{f(x) = (x - 1)(3x + 1)(x + 5i)(x - 5i)}
\]
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