To find the zeros of the polynomial [tex]\( f(x) = x^5 - 7x^4 + 20x^3 - 68x^2 + 99x - 45 \)[/tex], we will follow these steps:
1. Determine the possible rational zeros using the Rational Root Theorem.
2. Use synthetic division to test the possible zeros and factorize the polynomial.
3. Solve the resulting polynomial equations to find all the zeros.
### Step 1: Rational Root Theorem
The Rational Root Theorem states that any rational root of the polynomial [tex]\( f(x) \)[/tex] will be a fraction [tex]\( \frac{p}{q} \)[/tex], where:
- [tex]\( p \)[/tex] is a factor of the constant term (-45), and
- [tex]\( q \)[/tex] is a factor of the leading coefficient (1).
The constant term is -45 and the leading coefficient is 1. Therefore, the possible rational zeros are [tex]\( \pm 1, \pm 3, \pm 5, \pm 9, \pm 15, \pm 45 \)[/tex].
### Step 2: Synthetic Division
We will use synthetic division to test these possible zeros.
#### Testing [tex]\( x = 1 \)[/tex]:
[tex]\[
\begin{array}{r|rrrrrr}
1 & 1 & -7 & 20 & -68 & 99 & -45 \\
& & 1 & -6 & 14 & -54 & 45 \\
\hline
& 1 & -6 & 14 & -54 & 45 & 0 \\
\end{array}
\][/tex]
The remainder is 0, so [tex]\( x = 1 \)[/tex] is a root. Therefore, [tex]\( (x - 1) \)[/tex] is a factor.
Now, our polynomial can be written as:
[tex]\[ f(x) = (x - 1)(x^4 - 6x^3 + 14x^2 - 54x + 45) \][/tex]
#### Testing [tex]\( x = 1 \)[/tex] again for the quartic polynomial:
[tex]\[
\begin{array}{r|rrrrr}
1 & 1 & -6 & 14 & -54 & 45 \\
& & 1 & -5 & 9 & -45 \\
\hline
& 1 & -5 & 9 & -45 & 0 \\
\end{array}
\][/tex]
[tex]\( x = 1 \)[/tex] is again a root. Therefore, another factor is [tex]\( (x - 1) \)[/tex].
Now, our polynomial can be written as:
[tex]\[ f(x) = (x - 1)^2(x^3 - 5x^2 + 9x - 45) \][/tex]
#### Testing [tex]\( x = 1 \)[/tex] again for the cubic polynomial:
[tex]\[
\begin{array}{r|rrrr}
1 & 1 & -5 & 9 & -45 \\
& & 1 & -4 & 5 \\
\hline
& 1 & -4 & 5 & 0 \\
\end{array}
\][/tex]
[tex]\( x = 1 \)[/tex] is yet again a root. Therefore, another factor is [tex]\( (x - 1) \)[/tex].
We can write our polynomial as:
[tex]\[ f(x) = (x - 1)^3(x^2 - 4x + 5) \][/tex]
### Step 3: Solving the Quadratic Polynomial
We solve the quadratic polynomial [tex]\( x^2 - 4x + 5 \)[/tex] by using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = -4 \)[/tex], and [tex]\( c = 5 \)[/tex].
[tex]\[
x = \frac{4 \pm \sqrt{16 - 20}}{2} = \frac{4 \pm \sqrt{-4}}{2} = \frac{4 \pm 2i}{2} = 2 \pm i
\][/tex]
Therefore, [tex]\( x = 2 + i \)[/tex] and [tex]\( x = 2 - i \)[/tex] are the roots.
### Conclusion
The zeros of the polynomial [tex]\( f(x) = x^5 - 7x^4 + 20x^3 - 68x^2 + 99x - 45 \)[/tex], along with their multiplicities, are:
- [tex]\( x = 1 \)[/tex] (multiplicity 3)
- [tex]\( x = 2 + i \)[/tex] (multiplicity 1)
- [tex]\( x = 2 - i \)[/tex] (multiplicity 1)
Thus, the final answer is:
[tex]\[ \boxed{1 \text{ with multiplicity } 3, 2 + i, 2 - i} \][/tex]
