ind all complex zeros of the given polynomial​ function, and write the polynomial in completely factored form.

f left parenthesis x right parenthesis equals 4 x cubed minus 9 x squared minus 58 x plus 15f(x)=4x3−9x2−58x+15

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Part 1

Find the complex zeros of f. Repeat any zeros if their multiplicity is greater than 1.

xequals=enter your response here

To find the complex zeros of the polynomial function \( f(x) = 4x^3 - 9x^2 - 58x + 15 \), we will follow these steps:

### Step 1: Use the Rational Root Theorem
The Rational Root Theorem helps us identify possible rational zeros. The theorem suggests that any rational root of the polynomial must be a factor of the constant term (15) divided by a factor of the leading coefficient (4).

The factors of 15 are: \( \pm 1, \pm 3, \pm 5, \pm 15 \)  
The factors of 4 are: \( \pm 1, \pm 2, \pm 4 \)

Possible rational roots:  
\( \pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm 3, \pm \frac{3}{2}, \pm 5, \pm \frac{5}{2}, \pm 15, \pm \frac{15}{2} \)

### Step 2: Test potential rational roots using synthetic division
We will test a few values using synthetic division to see if we can find a root.

#### Testing \( x = 3 \) as a root:
Performing synthetic division of \( f(x) \) by \( (x - 3) \):

\[
\begin{array}{r|rrrr}
3 & 4 & -9 & -58 & 15 \\
  &   & 12 & 9  & -147 \\
\hline
  & 4 & 3  & -49 & -132 \\
\end{array}
\]

Since the remainder is not zero, \( x = 3 \) is **not** a root.

#### Testing \( x = -3 \):
Perform synthetic division again:

\[
\begin{array}{r|rrrr}
-3 & 4 & -9 & -58 & 15 \\
   &   & -12 & 63 & -15 \\
\hline
   & 4 & -21 & 5  & 0 \\
\end{array}
\]

The remainder is zero, so \( x = -3 \) is a root.

### Step 3: Divide the polynomial by \( (x + 3) \)
After dividing \( f(x) \) by \( (x + 3) \), the quotient is \( 4x^2 - 21x + 5 \).

### Step 4: Factor the quadratic \( 4x^2 - 21x + 5 \)
Now we factor the quadratic:

\[
4x^2 - 21x + 5 = (4x - 1)(x - 5)
\]

### Step 5: Write the completely factored form
The completely factored form of the polynomial is:

\[
f(x) = (x + 3)(4x - 1)(x - 5)
\]

### Step 6: Find the complex zeros
The zeros of the polynomial are obtained by solving \( (x + 3) = 0 \), \( (4x - 1) = 0 \), and \( (x - 5) = 0 \):

1. \( x = -3 \)
2. \( 4x - 1 = 0 \Rightarrow x = \frac{1}{4} \)
3. \( x = 5 \)

So, the complex zeros are:  
\[
x = -3, \quad x = \frac{1}{4}, \quad x = 5
\]

Since all the roots are real, there are no complex (non-real) zeros in this case.

Would you like more details or have any questions?

Here are 5 related questions to deepen your understanding:

1. How do you apply synthetic division to find polynomial roots?
2. What is the Rational Root Theorem, and how does it help in factoring polynomials?
3. Can all cubic polynomials be factored into real numbers only?
4. How do you determine the multiplicity of a zero in a polynomial?
5. How does factoring help in solving higher-degree polynomials?

**Tip:** Always check for rational roots first when factoring polynomials; it simplifies the process dramatically!

Find all complex zeros of the given polynomial function, and write the polynomial in completely factored form: f(x) = 4x^3 - 9x^2 - 58x + 15.

Explore how to find the complex zeros of the cubic polynomial f(x) = 4x^3 - 9x^2 - 58x + 15. This problem walks through the Rational Root Theorem, synthetic division, and factoring methods to solve the equation. Suitable for students in Grades 10-12.

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