a rectangle has an area of ( k^{2}+19k + 60 ) square inches. if the value of ( k ) and the dimensions of the…

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

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

Answer

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). 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 use 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 valid factor.
  • For the option "The width of the rectangle is (k + 4) inches": Since (k^{2}+19k + 60=(k + 15)(k + 4)), if we assume width (w=k + 4) and length (l=k + 15) (because area (A=l\times w) and (k) is a natural number), this is a valid factorization.
  • For the option "The length of the rectangle is (k-20) inches": If we use 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 valid factor.
  • For the option "The width of the rectangle is (k + 10) inches": If we use polynomial long - division (\frac{k^{2}+19k + 60}{k + 10}=\frac{k^{2}+10k+9k+90 - 30}{k + 10}=k+9-\frac{30}{k + 10}). Since (k) is a natural number, this is not a valid factor.

Answer:

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