Solved Polynomial Functions Worksheet $4.1 POLYNOMIAL | Chegg.com - Free Printable

Let’s solve each part of the problem step by step.

---

## Problem 1:
> Zeros: -3, 0, 4; degree 3

✔ Solution:

If a polynomial has real zeros at \( x = -3 \), \( x = 0 \), and \( x = 4 \), then it must have factors corresponding to those roots:

\[
(x + 3), \quad (x - 0) = x, \quad (x - 4)
\]

Since the degree is 3 and we have 3 distinct real zeros, we can write the polynomial as:

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

where \( a \) is a nonzero constant (leading coefficient). Since no specific point or leading coefficient is given, we can choose \( a = 1 \) for simplicity.

So,

\[
\boxed{f(x) = x(x + 3)(x - 4)}
\]

You can expand this if needed:

\[
f(x) = x[(x + 3)(x - 4)] = x[x^2 - 4x + 3x - 12] = x(x^2 - x - 12) = x^3 - x^2 - 12x
\]

✔ Final Answer: \( \boxed{f(x) = x^3 - x^2 - 12x} \)

---

## Problem 2:
> Zeros: -2 (multiplicity 2), 4 (multiplicity 1); degree 3

✔ Solution:

Multiplicity means how many times a factor appears.

- Zero at \( x = -2 \) with multiplicity 2 → factor: \( (x + 2)^2 \)
- Zero at \( x = 4 \) with multiplicity 1 → factor: \( (x - 4) \)

Total degree: \( 2 + 1 = 3 \), which matches.

So the polynomial is:

\[
f(x) = a(x + 2)^2(x - 4)
\]

Again, since no specific value is given, we can let \( a = 1 \).

So,

\[
\boxed{f(x) = (x + 2)^2(x - 4)}
\]

Expand if needed:

First, \( (x + 2)^2 = x^2 + 4x + 4 \)

Then multiply by \( (x - 4) \):

\[
(x^2 + 4x + 4)(x - 4) = x^3 - 4x^2 + 4x^2 - 16x + 4x - 16 = x^3 - 12x - 16
\]

✔ Final Answer: \( \boxed{f(x) = x^3 - 12x - 16} \)

---

## Problem 3:
> Construct a polynomial function that might have the given graph.

The graph shows:

- The curve crosses the x-axis at three points: approximately \( x = -2 \), \( x = 0 \), and \( x = 2 \).
- It passes through the point \( (3, 8) \).
- The ends go downward on both sides → even degree with negative leading coefficient.
- Since it crosses the x-axis 3 times, and the graph looks like a cubic (but wait — ends go down on both sides? That suggests even degree... but 3 real roots? Hmm.)

Wait — let’s analyze carefully.

Looking at the graph:

- It starts from bottom left (as \( x \to -\infty \), \( y \to -\infty \))
- Ends at bottom right (as \( x \to \infty \), \( y \to -\infty \))
→ So degree is even, and leading coefficient is negative.

But it crosses the x-axis at 3 points: likely \( x = -2 \), \( x = 0 \), \( x = 2 \). But if it’s even degree, it must have an even number of real roots counting multiplicity.

Possibility: one of the roots has multiplicity 2.

Looking at the graph near \( x = 0 \): the curve touches the x-axis and turns around — that’s characteristic of a double root (multiplicity 2).

At \( x = -2 \) and \( x = 2 \), it crosses the axis — so simple roots (multiplicity 1).

So total multiplicities: \( 1 + 2 + 1 = 4 \) → degree 4, even → matches end behavior.

So we assume:

- Root at \( x = -2 \) → factor: \( (x + 2) \)
- Root at \( x = 0 \) with multiplicity 2 → factor: \( x^2 \)
- Root at \( x = 2 \) → factor: \( (x - 2) \)

So polynomial form:

\[
f(x) = a(x + 2)(x^2)(x - 2)
\]

Simplify:

\[
f(x) = a x^2 (x + 2)(x - 2) = a x^2 (x^2 - 4) = a(x^4 - 4x^2)
\]

Now use the point \( (3, 8) \) to find \( a \):

Plug in \( x = 3 \), \( f(3) = 8 \):

\[
8 = a(3^4 - 4 \cdot 3^2) = a(81 - 36) = a(45)
\Rightarrow a = \frac{8}{45}
\]

✔ So the polynomial is:

\[
\boxed{f(x) = \frac{8}{45}(x^4 - 4x^2)}
\]

Or expanded:

\[
\boxed{f(x) = \frac{8}{45}x^4 - \frac{32}{45}x^2}
\]

---

## ✔ Final Answers:

1. \( \boxed{f(x) = x^3 - x^2 - 12x} \)

2. \( \boxed{f(x) = x^3 - 12x - 16} \)

3. \( \boxed{f(x) = \frac{8}{45}x^4 - \frac{32}{45}x^2} \)

---

Let me know if you want to see graphs or further simplifications!