Sure! Let's tackle this step-by-step.
### Given:
[tex]\[ f(x) = 9x^3 - 97x^2 + 304x - 182 \][/tex]
A known zero is [tex]\(5 + i\)[/tex].
### Step 1: Identify the Conjugate Zero
Since the coefficients of the polynomial are real, the complex conjugate of any complex zero is also a zero. Therefore, if [tex]\(5 + i\)[/tex] is a zero, [tex]\(5 - i\)[/tex] must also be a zero.
### Step 2: Form the Quadratic Factor
The zeros [tex]\(5 + i\)[/tex] and [tex]\(5 - i\)[/tex] imply a quadratic factor of the polynomial [tex]\( f(x) \)[/tex]:
[tex]\[
(x - (5 + i))(x - (5 - i))
\][/tex]
This product simplifies to:
[tex]\[
(x - 5 - i)(x - 5 + i) = (x - 5)^2 - i^2 = (x - 5)^2 + 1
\][/tex]
[tex]\[
= x^2 - 10x + 25 + 1 = x^2 - 10x + 26
\][/tex]
### Step 3: Perform Polynomial Division
We need to divide the polynomial [tex]\(f(x)\)[/tex] by the quadratic factor we just found:
[tex]\[
\frac{9x^3 - 97x^2 + 304x - 182}{x^2 - 10x + 26}
\][/tex]
The quotient from this division is a linear polynomial (since a cubic divided by a quadratic leaves a linear), which we can form as:
[tex]\[
qx + r = 9x + p
\][/tex]
In this case, we already know that the quotient is of the form [tex]\(9(x - 7/9)\)[/tex].
### Step 4: Find the Remaining Zero
Since:
[tex]\[
9x^3 - 97x^2 + 304x - 182 = (x^2 - 10x + 26)(9x - 7)
\][/tex]
Equating the remaining factor to zero gives us:
[tex]\[
9x - 7 = 0 \Rightarrow x = \frac{7}{9}
\][/tex]
### Step 5: List of All Zeros
The zeros of the polynomial are:
[tex]\[
5 + i, 5 - i, \frac{7}{9}
\][/tex]
### Step 6: Factor the Polynomial as Product of Linear Factors
Given the zeros, the polynomial can be factored as:
[tex]\[
f(x) = 9(x - (5+i))(x - (5-i))(x - \frac{7}{9})
\][/tex]
We can write the polynomial in its completely factored form:
[tex]\[
f(x) = 9 \left( x - 5 - i \right) \left( x - 5 + i \right) \left( x - \frac{7}{9} \right)
\][/tex]
### Step 7: Simplify the Factors
[tex]\[
= 9[(x - 5)^2 + 1]\left( x - \frac{7}{9} \right)
\][/tex]
### Step 8: Solve the Equation [tex]\( f(x) = 0 \)[/tex]
To solve [tex]\( f(x) = 0 \)[/tex], we set each factor to zero:
[tex]\[
(x - (5+i)) = 0 \Rightarrow x = 5 + i
\][/tex]
[tex]\[
(x - (5-i)) = 0 \Rightarrow x = 5 - i
\][/tex]
[tex]\[
(x - \frac{7}{9}) = 0 \Rightarrow x = \frac{7}{9}
\][/tex]
### Summary
(a) All the zeros are: [tex]\( 5 + i \)[/tex], [tex]\( 5 - i \)[/tex], [tex]\(\frac{7}{9}\)[/tex].
(b) The polynomial factored as a product of linear factors is:
[tex]\[
f(x) = 9\left((x - (5 + i))(x - (5 - i))(x - \frac{7}{9})\right)
\][/tex]
(c) The solutions to the equation [tex]\( f(x) = 0 \)[/tex] are:
[tex]\[
x = 5 + i, x = 5 - i, x = \frac{7}{9}
\][/tex]
