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 area expression

We need to factor the quadratic expression \(k^2 + 19k + 60\). We look for 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, \(k^2+19k + 60=(k + 15)(k + 4)\)? Wait, no, wait: Wait, \(15+4 = 19\)? Wait, \(15 + 4=19\)? Wait, \(15\times4 = 60\), yes. Wait, but let's check the options. Wait, maybe I made a mistake. Wait, the options have \(k + 10\), \(k + 4\), \(k-5\), \(k - 20\). Wait, maybe I miscalculated. Wait, let's try again. Let's factor \(k^2+19k + 60\). Let's find two numbers \(a\) and \(b\) such that \(a\times b=60\) and \(a + b = 19\). Wait, \(15\) and \(4\) give \(15\times4 = 60\) and \(15+4 = 19\), so \(k^2+19k + 60=(k + 15)(k + 4)\)? But the options have \(k + 10\). Wait, maybe I made a mistake. Wait, maybe the quadratic is \(k^2+19k + 60\)? Wait, no, maybe the quadratic is \(k^2+19k + 60\). Wait, let's check the options. The options for width or length are \(k - 5\), \(k + 4\), \(k - 20\), \(k + 10\). Wait, maybe I factored wrong. Wait, let's try again. Let's suppose the quadratic is \(k^2+19k + 60\). Wait, maybe the correct factors are \((k + 15)(k + 4)\) is wrong. Wait, no, \(15+4=19\), \(15\times4 = 60\). But the options have \(k + 10\). Wait, maybe the quadratic is \(k^2+19k + 60\) is a typo? Or maybe I misread. Wait, the problem says "a rectangle has an area of \(k^2 + 19k + 60\) square inches". Wait, let's check the options. Let's test each option.

Let's take the option "The width of the rectangle is \(k + 4\) inches". Let's assume length is \(k + 15\) (since \((k + 15)(k + 4)=k^2+19k + 60\)). Wait, but the options have \(k + 4\) as width. Let's check other options.

Option 1: Length \(k - 5\). If length is \(k - 5\), then width would be \(\frac{k^2+19k + 60}{k - 5}\). Let's perform polynomial division or check if \(k - 5\) is a factor. Using Remainder Theorem, substitute \(k = 5\) into \(k^2+19k + 60\): \(25+95 + 60=180
eq0\), so not a factor.

Option 2: Width \(k + 4\). If width is \(k + 4\), then length is \(\frac{k^2+19k + 60}{k + 4}\). Let's divide \(k^2+19k + 60\) by \(k + 4\). Using polynomial long division: \(k^2+19k + 60=(k + 4)(k + 15)+0\)? Wait, \( (k + 4)(k + 15)=k^2+19k + 60\), yes. Wait, but the options have \(k + 4\) as width. Let's check the other options.

Option 3: Length \(k - 20\). Substitute \(k = 20\) into \(k^2+19k + 60\): \(400+380 + 60=840
eq0\), so not a factor.

Option 4: Width \(k + 10\). Divide \(k^2+19k + 60\) by \(k + 10\). Using Remainder Theorem, substitute \(k=-10\): \(100-190 + 60=-30
eq0\), so not a factor.

Wait, so the correct factoring is \((k + 15)(k + 4)\), so if width is \(k + 4\), then length is \(k + 15\) (which is a natural number as \(k\) is natural). So the option "The width of the rectangle is \(k + 4\) inches" is correct.

Step2: Verify each option

• For "The length of the rectangle is \(k - 5\) inches": As we saw, \(k - 5\) is not a factor of \(k^2+19k + 60\) (remainder non - zero when \(k = 5\)).
• For "The width of the rectangle is \(k + 4\) inches": Since \(k^2+19k + 60=(k + 15)(k + 4)\), and both \(k+15\) and \(k + 4\) are natural numbers (as \(k\) is natural), this is a valid dimension.
• For "The length of the rectangle is \(k - 20\) inches": \(k-20\) is not a factor (remainder non - zero when \(k = 20\)).
• For "The width of the rectangle is \(k + 10\) inches": \(k + 10\) is not a factor (remainder non - zero when \(k=-10\)).

Answer:

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