Question

question 1-14
the rectangular area of the new school courtyard is represented by the expression: $4c^2 - 36$. which expression could represent the length times the width of the courtyard?
$\bigcirc\\ (2c - 6)^2$
$\bigcirc\\ (2c - 6)(2c + 6)$
$\bigcirc\\ 4(c^2 - 32)$
$\bigcirc\\ (4c - 6)(c + 6)$

Explanation:

Step1: Factor out the common factor

First, factor out the common factor of 4 from the expression \(4c^{2}-36\). We get \(4(c^{2} - 9)\).

Step2: Apply the difference of squares formula

The expression \(c^{2}-9\) is a difference of squares, which can be factored using the formula \(a^{2}-b^{2}=(a - b)(a + b)\). Here, \(a = c\) and \(b = 3\), so \(c^{2}-9=(c - 3)(c + 3)\). But we can also rewrite \(4c^{2}-36\) as \((2c)^{2}-6^{2}\), which is also a difference of squares. Using the difference of squares formula \(a^{2}-b^{2}=(a - b)(a + b)\) where \(a = 2c\) and \(b = 6\), we have \((2c)^{2}-6^{2}=(2c - 6)(2c + 6)\).

Step3: Analyze other options

• Option 1: \((2c - 6)^{2}=(2c - 6)(2c - 6)\), which is not equal to \(4c^{2}-36\) because \((2c - 6)(2c - 6)=4c^{2}-24c + 36

eq4c^{2}-36\).

• Option 3: \(4(c^{2}-32)=4c^{2}-128

eq4c^{2}-36\).

• Option 4: \((4c - 6)(c + 6)=4c^{2}+24c-6c - 36=4c^{2}+18c - 36

eq4c^{2}-36\).

Answer:

\((2c - 6)(2c + 6)\) (the second option)