Question

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 \(a\) and \(b\) such that \(a + b=19\) and \(a\times b = 60\).
The numbers are \(15\) and \(4\) since \(15+4 = 19\) and \(15\times4=60\).
So, \(k^{2}+19k + 60=(k + 15)(k+4)\)

Step2: Analyze each option

• Option 1: If the length is \(k - 5\), then \((k - 5)\times w=k^{2}+19k + 60\). Using polynomial long - division \(\frac{k^{2}+19k + 60}{k - 5}=\frac{k^{2}-5k+24k-120 + 180}{k - 5}=k + 24+\frac{180}{k - 5}\). Since \(k\) is a natural number, this is not a polynomial with natural number coefficients for the width.
• Option 2: If the width is \(k + 4\), then the length is \(\frac{k^{2}+19k + 60}{k + 4}\). Using polynomial long - division \(\frac{k^{2}+19k+60}{k + 4}=\frac{k^{2}+4k+15k + 60}{k + 4}=k + 15\). Both \(k + 15\) and \(k + 4\) are polynomials with natural number coefficients when \(k\) is a natural number.
• Option 3: If the length is \(k-20\), then \((k - 20)\times w=k^{2}+19k + 60\). Using polynomial long - division \(\frac{k^{2}+19k + 60}{k - 20}=\frac{k^{2}-20k+39k-780+840}{k - 20}=k + 39+\frac{840}{k - 20}\). Since \(k\) is a natural number, this is not a polynomial with natural number coefficients for the width.
• Option 4: If the width is \(k + 10\), then the length is \(\frac{k^{2}+19k + 60}{k + 10}\). Using polynomial long - division \(\frac{k^{2}+19k+60}{k + 10}=\frac{k^{2}+10k+9k + 60}{k + 10}=k + 9+\frac{- 30}{k + 10}\). Since \(k\) is a natural number, this is not a polynomial with natural number coefficients for the length.

Answer:

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