Answer :
To determine which value of \(c\) makes the expression \(8x + cy\) completely factored, we need to identify a \(c\) such that the expression can be written as a product of simpler expressions.
Consider the expression \(8x + cy\). For this expression to be completely factored, \(c\) should allow us to factor out a common term from both \(8x\) and \(cy\).
We know that \(8x\) can be factored as \(8 \cdot x\). Thus, \(cy\) must be able to similarly factor into \(8 \cdot (\text{some term}) \cdot y\) for a common factor to be present.
To achieve this, we need \(cy = 8(\text{some integer})\). This implies that \(c\) must be a multiple of 8 to factor out the 8 from the expression.
Given the values \(2\), \(7\), \(12\), and \(16\):
- When \(c = 2\), \(8x + 2y\) cannot be factored to include an 8.
- When \(c = 7\), \(8x + 7y\) cannot be factored to include an 8.
- When \(c = 12\), \(8x + 12y\) cannot be factored to include an 8.
- When \(c = 16\), the expression becomes \(8x + 16y\), and it can be factored out as \(8(x + 2y)\) because both \(8x\) and \(16y\) have a common factor of 8.
Therefore, the value of [tex]\(c\)[/tex] which allows [tex]\(8x + cy\)[/tex] to be completely factored is [tex]\( \boxed{16} \)[/tex].
Consider the expression \(8x + cy\). For this expression to be completely factored, \(c\) should allow us to factor out a common term from both \(8x\) and \(cy\).
We know that \(8x\) can be factored as \(8 \cdot x\). Thus, \(cy\) must be able to similarly factor into \(8 \cdot (\text{some term}) \cdot y\) for a common factor to be present.
To achieve this, we need \(cy = 8(\text{some integer})\). This implies that \(c\) must be a multiple of 8 to factor out the 8 from the expression.
Given the values \(2\), \(7\), \(12\), and \(16\):
- When \(c = 2\), \(8x + 2y\) cannot be factored to include an 8.
- When \(c = 7\), \(8x + 7y\) cannot be factored to include an 8.
- When \(c = 12\), \(8x + 12y\) cannot be factored to include an 8.
- When \(c = 16\), the expression becomes \(8x + 16y\), and it can be factored out as \(8(x + 2y)\) because both \(8x\) and \(16y\) have a common factor of 8.
Therefore, the value of [tex]\(c\)[/tex] which allows [tex]\(8x + cy\)[/tex] to be completely factored is [tex]\( \boxed{16} \)[/tex].
