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

We factor \(k^{2}+19k + 60\). We need to find two numbers that multiply to \(60\) and add up to \(19\). The numbers are \(15\) and \(4\). So, \(k^{2}+19k + 60=(k + 15)(k+4)\)

Step2: Analyze each option

• For the option "The length of the rectangle is \(k - 5\) inches": If \((k - 5)\) is a factor, then when we divide \(k^{2}+19k + 60\) by \((k - 5)\) (using polynomial long - division or synthetic division), the remainder is non - zero.
• For the option "The width of the rectangle is \(k + 4\) inches": Since \(k^{2}+19k + 60=(k + 15)(k + 4)\), if we assume the width is \(k + 4\) and the length is \(k + 15\) (because area of a rectangle \(A=\text{length}\times\text{width}\)), and \(k\) is a natural number, then both \(k + 4\) and \(k + 15\) are natural numbers.
• For the option "The length of the rectangle is \(k - 20\) inches": If \((k - 20)\) is a factor, then when we divide \(k^{2}+19k + 60\) by \((k - 20)\) (using polynomial long - division or synthetic division), the remainder is non - zero.
• For the option "The width of the rectangle is \(k + 10\) inches": If \((k + 10)\) is a factor, then when we divide \(k^{2}+19k + 60\) by \((k + 10)\) (using polynomial long - division or synthetic division), the remainder is non - zero.

Answer:

The width of the rectangle is \(k + 4\) inches.