Question

a rectangle has an area of ( k^2 + 19k + 60 ) square inches. if the value of ( k ) and the dimensions of the rectangle are all natural numbers, which statement about the rectangle could be true?

• the length of the rectangle is ( k - 5 ) inches.
• the width of the rectangle is ( k + 4 ) inches.
• the length of the rectangle is ( k - 20 ) inches.
• the width of the rectangle is ( k + 10 ) inches.

Explanation:

Step1: Factor the quadratic expression

To find the dimensions of the rectangle, we factor the area expression \(k^2 + 19k + 60\). We need two numbers that multiply to \(60\) and add up to \(19\). The numbers are \(15\) and \(4\) (since \(15\times4 = 60\) and \(15 + 4=19\)). So, \(k^2 + 19k + 60=(k + 15)(k + 4)\).

Step2: Analyze each option

• For the first option: The length is \(k - 5\). But our factored form has terms with \(+\) signs, and \(k-5\) doesn't match the factored factors, so this is false.
• For the second option: The width is \(k + 4\). Since the area factors to \((k + 15)(k + 4)\), and \(k\) is a natural number, \(k + 4\) can be a valid width (as dimensions are natural numbers, \(k+4>0\) for natural \(k\)).
• For the third option: The length is \(k - 20\). Similar to the first option, the factored form has \(+\) signs, and \(k - 20\) would be negative for small natural \(k\) (e.g., \(k = 1\), \(1-20=-19\)), so this is false.
• For the fourth option: The width is \(k + 10\). Our factored form is \((k + 15)(k + 4)\), and \(k + 10\) is not one of the factors, so this is false.

Answer:

The width of the rectangle is \(k + 4\) inches. (Corresponding to the second option in the original multiple - choice question)