Degree 4;    zeros -4, 0, 1, 6;    coefficient of x^(3)is 6

Given that the polynomial has degree 4 and the zeros are \(-4\), \(0\), \(1\), and \(6\), we can construct the polynomial based on these roots.

The general form of a polynomial with these roots is:

\[
P(x) = a(x + 4)(x)(x - 1)(x - 6)
\]

Where \(a\) is a constant that needs to be determined, and \(x = 0\) is the root from which we infer that one factor is simply \(x\).

### Step 1: Expand the polynomial
First, let's expand \(P(x)\) to express it fully:

\[
P(x) = a \cdot x(x + 4)(x - 1)(x - 6)
\]

We can group the terms for easier expansion:
- First, expand \((x + 4)(x - 1)\):
  \[
  (x + 4)(x - 1) = x^2 - x + 4x - 4 = x^2 + 3x - 4
  \]
- Then expand \((x^2 + 3x - 4)(x - 6)\):
  \[
  (x^2 + 3x - 4)(x - 6) = x^3 - 6x^2 + 3x^2 - 18x - 4x + 24 = x^3 - 3x^2 - 22x + 24
  \]

So, the expanded form is:
\[
P(x) = a \cdot x(x^3 - 3x^2 - 22x + 24)
\]
Multiplying by \(x\):
\[
P(x) = a(x^4 - 3x^3 - 22x^2 + 24x)
\]

### Step 2: Determine the coefficient of \(x^3\)
From the equation above, the coefficient of \(x^3\) is \(-3a\). According to the problem, the coefficient of \(x^3\) is 6. Therefore, we set:
\[
-3a = 6
\]

Solving for \(a\):
\[
a = -2
\]

### Step 3: Final polynomial
Substitute \(a = -2\) back into the polynomial:
\[
P(x) = -2(x^4 - 3x^3 - 22x^2 + 24x)
\]

Simplifying:
\[
P(x) = -2x^4 + 6x^3 + 44x^2 - 48x
\]

Thus, the polynomial is:
\[
P(x) = -2x^4 + 6x^3 + 44x^2 - 48x
\]

Would you like any further details or clarifications? Here are 5 related questions to explore further:

1. How would the polynomial change if the coefficient of \(x^2\) were specified?
2. What is the sum of the roots of the polynomial?
3. How would you find the y-intercept of this polynomial?
4. How would the graph of this polynomial look based on its roots and leading coefficient?
5. What would the effect of multiplying all coefficients by a constant be on the polynomial?

**Tip:** Always check your work by substituting the known roots back into the polynomial to verify the solution!

Degree 4; zeros -4, 0, 1, 6; coefficient of x^(3) is 6

Discover how to construct and solve a degree 4 polynomial with given zeros -4, 0, 1, 6 and a specified x^3 coefficient of 6. This detailed step-by-step solution covers polynomial expansion and coefficient matching, suitable for high school students in Grades 10-12.

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