( $$ f(x)=7 x^{4}-6 x^{3}+174 x^{2}-150 x-25 $$ a) Find the rational zeros of $$ f(x) $$, and then the other zeros. b) Factor $$ f(x) $$ into linear factors. a) Find the rational zeros of $$ f(x) $$. Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. A. The rational zeros are $$ 1,-\frac{1}{7} $$. (Use a comma to separate answers as needed. Express complex numbers in terms of $$ i $$.) B. There are no rational zeros. Find the other zeros of $$ f(x) $$. Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. A. The other zeros are (Use a comma to separate answers as needed. Express complex numbers in terms of $$ i $$.) B. There are no other zeros.

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We start with the polynomial
$$
f(x)=7x^4-6x^3+174x^2-150x-25.
$$

**Step 1. Find the Rational Zeros**

The Rational Root Theorem tells us that any rational zero must be of the form
$$
\frac{p}{q},
$$
where $$ p $$ divides the constant term $$-25$$ and $$ q $$ divides the leading coefficient $$7$$. Thus, the possible values for $$ p $$ are $$\pm1, \pm5, \pm25$$ and for $$ q $$ are $$\pm1, \pm7$$. This gives the possible rational roots:
$$
\pm1, \pm5, \pm25, \pm\frac{1}{7}, \pm\frac{5}{7}, \pm\frac{25}{7}.
$$

Test $$ x=1 $$:
$$
f(1)=7(1)^4-6(1)^3+174(1)^2-150(1)-25=7-6+174-150-25=0.
$$
Since $$ f(1)=0 $$, we have a rational zero at $$ x=1 $$.

Now, perform synthetic division to factor out $$ x-1 $$ from $$ f(x) $$. Using the coefficients $$ 7, -6, 174, -150, -25 $$:

$$
\begin{array}{r|rrrrr}
1 & 7 & -6 & 174 & -150 & -25 \ \hline
  &   & 7  & 1    & 175  & 25 \
  & 7 & 1  & 175  & 25   & 0 \
\end{array}
$$
The quotient is $$ 7x^3+x^2+175x+25 $$.

Next, test a potential rational root for the cubic $$ 7x^3+x^2+175x+25 $$. Consider $$ x=-\frac{1}{7} $$:

$$
7\left(-\frac{1}{7}\right)^3+\left(-\frac{1}{7}\right)^2+175\left(-\frac{1}{7}\right)+25.
$$
Compute each term:
$$
\left(-\frac{1}{7}\right)^3=-\frac{1}{343} \quad \Rightarrow \quad 7\left(-\frac{1}{343}\right)=-\frac{7}{343}=-\frac{1}{49},
$$
$$
\left(-\frac{1}{7}\right)^2=\frac{1}{49},
$$
$$
175\left(-\frac{1}{7}\right)=-25.
$$
Adding up:
$$
-\frac{1}{49}+\frac{1}{49}-25+25=0.
$$
Thus, $$ x=-\frac{1}{7} $$ is also a root.

Use synthetic division on $$ 7x^3+x^2+175x+25 $$ with $$ x = -\frac{1}{7} $$:
$$
\begin{array}{r|rrrr}
-\frac{1}{7} & 7 & 1 & 175 & 25 \ \hline
             &   & -1 & 0   & -25 \
             & 7 & 0 & 175 & 0 \
\end{array}
$$
The depressed polynomial is $$ 7x^2+175 $$, which factors as:
$$
7(x^2+25).
$$

Thus, the factorization so far is:
$$
f(x)=(x-1)\left(x+\frac{1}{7}\right)7(x^2+25).
$$

To have linear factors, note that:
$$
x+\frac{1}{7}=\frac{1}{7}(7x+1).
$$
Therefore, rewriting the full factorization:
$$
f(x)=7(x-1)\left(\frac{1}{7}(7x+1)\right)(x^2+25)=(x-1)(7x+1)(x^2+25).
$$

**Step 2. Find the Other Zeros**

We already found the rational zeros: $$ x=1 $$ and $$ x=-\frac{1}{7} $$. The remaining factor is $$ x^2+25 $$. Set it equal to zero:
$$
x^2+25=0 \quad \Rightarrow \quad x^2=-25 \quad \Rightarrow \quad x=\pm\sqrt{-25}=\pm5i.
$$

**Step 3. Express the Answer**

a) The rational zeros of $$ f(x) $$ are $$ 1 $$ and $$-\frac{1}{7}$$.

b) The other zeros of $$ f(x) $$ are $$ 5i $$ and $$-5i$$.

The complete factorization of $$ f(x) $$ into linear factors is:
$$
f(x)=(x-1)(7x+1)(x-5i)(x+5i).
$$
a) The rational zeros of $$ f(x) $$ are $$ 1 $$ and $$-\frac{1}{7}$$. b) The other zeros of $$ f(x) $$ are $$ 5i $$ and $$-5i$$. The complete factorization of $$ f(x) $$ into linear factors is: $$ f(x)=(x-1)(7x+1)(x-5i)(x+5i). $$

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