Question

1. the length of a rectangle is represented by ( x^2 + 3x + 2 ), and the width is represented by ( 4x ). express the perimeter of the rectangle as a trinomial. express the area of the rectangle as a trinomial.

Explanation:

Step1: Recall Perimeter Formula

Perimeter of rectangle: \( P = 2(l + w) \), where \( l \) is length, \( w \) is width.
Given \( l = x^2 + 3x + 2 \), \( w = 4x \).
Substitute into formula:
\( P = 2[(x^2 + 3x + 2) + 4x] \)

Step2: Simplify Perimeter Expression

Combine like terms inside the brackets:
\( (x^2 + 3x + 2) + 4x = x^2 + (3x + 4x) + 2 = x^2 + 7x + 2 \)
Multiply by 2:
\( P = 2(x^2 + 7x + 2) = 2x^2 + 14x + 4 \)

Step3: Recall Area Formula

Area of rectangle: \( A = l \times w \).
Substitute \( l = x^2 + 3x + 2 \), \( w = 4x \):
\( A = (x^2 + 3x + 2)(4x) \)

Step4: Simplify Area Expression

Distribute \( 4x \) to each term:
\( 4x \cdot x^2 + 4x \cdot 3x + 4x \cdot 2 = 4x^3 + 12x^2 + 8x \) (Wait, the problem says "trinomial", but this is a cubic. Wait, maybe a typo? Wait, length is \( x^2 + 3x + 2 \), width is \( 4x \). Wait, area is \( (x^2 + 3x + 2)(4x) = 4x^3 + 12x^2 + 8x \), but perimeter is \( 2x^2 + 14x + 4 \) (trinomial). Wait, maybe the width is \( 4 \) (not \( 4x \))? But the problem says \( 4x \). Wait, let's check again.

Wait, the problem says: "Express the perimeter of the rectangle as a trinomial. Express the area of the rectangle as a trinomial."
Wait, if width is \( 4x \), area is \( (x^2 + 3x + 2)(4x) = 4x^3 + 12x^2 + 8x \) (cubic, 3 terms, so trinomial? A trinomial is a polynomial with 3 terms, regardless of degree. So yes, 4x³ +12x² +8x is a trinomial. Perimeter: 2(x² +3x +2 +4x) = 2(x² +7x +2) = 2x² +14x +4 (trinomial).

Answer:

Perimeter: \( \boldsymbol{2x^2 + 14x + 4} \)
Area: \( \boldsymbol{4x^3 + 12x^2 + 8x} \)