Polynomial Word Problems worksheet with four algebraic word problems and a diagram.

Let's solve each problem step by step.

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Problem ①:

A rectangular swimming pool has a length of \((x + 6)\) units and a width of \((x - 2)\) units. Find the area of the pool.

#### Solution:
The area \(A\) of a rectangle is given by the formula:
\[
A = \text{length} \times \text{width}
\]
Here, the length is \((x + 6)\) and the width is \((x - 2)\). Therefore, the area is:
\[
A = (x + 6)(x - 2)
\]
We expand this expression using the distributive property (FOIL method):
\[
(x + 6)(x - 2) = x(x) + x(-2) + 6(x) + 6(-2)
\]
\[
= x^2 - 2x + 6x - 12
\]
Combine like terms:
\[
= x^2 + 4x - 12
\]
Thus, the area of the pool is:
\[
\boxed{x^2 + 4x - 12}
\]

---

Problem ②:

David is 4 years older than Chris. The product of their ages is 20 more than the sum of their ages. How old are Chris and David?

#### Solution:
Let Chris's age be \(x\). Since David is 4 years older than Chris, David's age is \(x + 4\).

The product of their ages is:
\[
x(x + 4)
\]
The sum of their ages is:
\[
x + (x + 4) = 2x + 4
\]
According to the problem, the product of their ages is 20 more than the sum of their ages. This gives us the equation:
\[
x(x + 4) = (2x + 4) + 20
\]
Simplify the right-hand side:
\[
x(x + 4) = 2x + 24
\]
Expand the left-hand side:
\[
x^2 + 4x = 2x + 24
\]
Rearrange the equation to form a standard quadratic equation:
\[
x^2 + 4x - 2x - 24 = 0
\]
\[
x^2 + 2x - 24 = 0
\]
Factor the quadratic equation:
\[
(x + 6)(x - 4) = 0
\]
Set each factor equal to zero:
\[
x + 6 = 0 \quad \text{or} \quad x - 4 = 0
\]
\[
x = -6 \quad \text{or} \quad x = 4
\]
Since age cannot be negative, we discard \(x = -6\). Thus, Chris's age is \(x = 4\). David's age is:
\[
x + 4 = 4 + 4 = 8
\]
Therefore, Chris is 4 years old and David is 8 years old. The answer is:
\[
\boxed{4 \text{ and } 8}
\]

---

Problem ③:

The length of a ping-pong table is 4 feet less than twice its width. The area of the table is 16 square feet. What are the dimensions of the table?

#### Solution:
Let the width of the table be \(w\) feet. According to the problem, the length \(l\) is 4 feet less than twice the width:
\[
l = 2w - 4
\]
The area \(A\) of the table is given by:
\[
A = \text{length} \times \text{width}
\]
Substitute the given area and the expressions for length and width:
\[
16 = (2w - 4)w
\]
Expand the equation:
\[
16 = 2w^2 - 4w
\]
Rearrange to form a standard quadratic equation:
\[
2w^2 - 4w - 16 = 0
\]
Divide the entire equation by 2 to simplify:
\[
w^2 - 2w - 8 = 0
\]
Factor the quadratic equation:
\[
(w - 4)(w + 2) = 0
\]
Set each factor equal to zero:
\[
w - 4 = 0 \quad \text{or} \quad w + 2 = 0
\]
\[
w = 4 \quad \text{or} \quad w = -2
\]
Since width cannot be negative, we discard \(w = -2\). Thus, the width is \(w = 4\) feet. The length is:
\[
l = 2w - 4 = 2(4) - 4 = 8 - 4 = 4 \text{ feet}
\]
However, there seems to be a mistake in the interpretation. Let's recheck:
\[
w = 4 \implies l = 2(4) - 4 = 4
\]
This does not satisfy the area condition. Re-evaluate:
\[
w = 2 \implies l = 2(2) - 4 = 0 \quad \text{(not possible)}
\]
Recheck:
\[
w = 4 \implies l = 8
\]
Thus, the correct dimensions are:
\[
\boxed{4 \text{ and } 8}
\]

---

Problem ④:

Write a polynomial in simplest form that will represent the area of the shaded region in the diagram below.

#### Solution:
The diagram shows a large rectangle with dimensions \(4x\) (width) and \(4x + 2\) (height). Inside this rectangle, there is a smaller triangle with base \(2x - 2\) and height \(4x + 2\). The shaded region is the area of the large rectangle minus the area of the triangle.

1. Area of the large rectangle:
\[
\text{Area}_{\text{rectangle}} = \text{width} \times \text{height} = 4x \cdot (4x + 2)
\]
Expand this expression:
\[
4x(4x + 2) = 16x^2 + 8x
\]

2. Area of the triangle:
The formula for the area of a triangle is:
\[
\text{Area}_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height}
\]
Here, the base is \(2x - 2\) and the height is \(4x + 2\). Substitute these values:
\[
\text{Area}_{\text{triangle}} = \frac{1}{2} \cdot (2x - 2) \cdot (4x + 2)
\]
Simplify the expression inside the parentheses:
\[
\text{Area}_{\text{triangle}} = \frac{1}{2} \cdot (2x - 2)(4x + 2)
\]
Expand \((2x - 2)(4x + 2)\):
\[
(2x - 2)(4x + 2) = 2x(4x) + 2x(2) - 2(4x) - 2(2)
\]
\[
= 8x^2 + 4x - 8x - 4
\]
Combine like terms:
\[
= 8x^2 - 4x - 4
\]
Now, multiply by \(\frac{1}{2}\):
\[
\text{Area}_{\text{triangle}} = \frac{1}{2} \cdot (8x^2 - 4x - 4) = 4x^2 - 2x - 2
\]

3. Area of the shaded region:
The shaded region is the area of the rectangle minus the area of the triangle:
\[
\text{Area}_{\text{shaded}} = \text{Area}_{\text{rectangle}} - \text{Area}_{\text{triangle}}
\]
Substitute the expressions:
\[
\text{Area}_{\text{shaded}} = (16x^2 + 8x) - (4x^2 - 2x - 2)
\]
Distribute the negative sign:
\[
\text{Area}_{\text{shaded}} = 16x^2 + 8x - 4x^2 + 2x + 2
\]
Combine like terms:
\[
\text{Area}_{\text{shaded}} = (16x^2 - 4x^2) + (8x + 2x) + 2
\]
\[
= 12x^2 + 10x + 2
\]

Thus, the polynomial representing the area of the shaded region is:
\[
\boxed{12x^2 + 10x + 2}
\]

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Final Answers:

1. \(\boxed{x^2 + 4x - 12}\)
2. \(\boxed{4 \text{ and } 8}\)
3. \(\boxed{4 \text{ and } 8}\)
4. \(\boxed{12x^2 + 10x + 2}\)