Question

the diagram represents the factorization of $a^2 + 8a + 12$. what is the missing number that will complete the factorization? diagram with a table: first row has a and 3 (left cell empty), second row: a, $a^2$, 6a; third row: 2, 2a, 12 options: 6, 8, 12, 24

Explanation:

Step1: Recall factoring trinomials

To factor \(a^2 + 8a + 12\), we need two numbers that multiply to \(12\) (the constant term) and add up to \(8\) (the coefficient of \(a\)). Let the missing number be \(x\). We know that in the table, the product of the first column's second row (\(2\)) and the first row's second column (\(x\)) should be \(2a\)? Wait, no, looking at the table structure: the top row has \(a\) and \(x\), the left column has \(a\) and \(2\). Then the cells are: \(a\times a = a^2\), \(a\times x = 6a\) (wait, no, the middle cell of the second row is \(6a\)? Wait, no, the table is:

First row: [blank], \(a\), \(x\)

Second row: \(a\), \(a^2\), \(6a\)

Third row: \(2\), \(2a\), \(12\)

Wait, actually, the area model for factoring: the trinomial \(a^2 + 8a + 12\) can be represented as a rectangle with area \(a^2 + 8a + 12\), which is the sum of four smaller rectangles. The top row's second column is \(a\), third column is \(x\); left column's second row is \(a\), third row is \(2\). Then:

• The top - left blank cell times \(a\) should be \(a^2\)? No, wait, the second row first column is \(a\), second column is \(a^2\) (so \(a\times a=a^2\)), third column is \(6a\) (so \(a\times x = 6a\)? Wait, no, the third row third column is \(12\), which is \(2\times x\) (since third row first column is \(2\), third column is \(12\), so \(2\times x = 12\)? No, wait, the middle term is \(8a=6a + 2a\). Wait, let's use the fact that in the area model, the sum of the middle terms: \(6a+2a = 8a\). Also, the product of the two numbers (the ones we multiply to get the constant term) should be \(12\) and add to \(8\). We know one of the numbers is \(2\) (from the third row first column), so the other number \(x\) should satisfy \(2\times x=12\)? No, wait, the third row third column is \(12\), which is \(2\times x\)? Wait, no, the third row is \(2\), \(2a\) (which is \(2\times a\)), and \(12\) (which is \(2\times x\)). So \(2\times x = 12\)? No, that would make \(x = 6\), but let's check the middle term. The second row third column is \(6a\), which is \(a\times x\). So \(a\times x=6a\), dividing both sides by \(a\) (assuming \(a

eq0\)), we get \(x = 6\). Let's verify: the trinomial factored would be \((a + 2)(a + 6)\), because \(a\times a=a^2\), \(a\times6 = 6a\), \(2\times a=2a\), \(2\times6 = 12\). Then \(6a+2a = 8a\), so the trinomial is \(a^2+8a + 12=(a + 2)(a + 6)\). So in the top row, the third column should be \(6\), because the first factor is \((a + 6)\) (top row: \(a\) and \(6\)), and the second factor is \((a + 2)\) (left column: \(a\) and \(2\)).

Step2: Check the product of the missing number and 2

Wait, the third row third cell is \(12\), which is \(2\times6 = 12\), and the second row third cell is \(6a\), which is \(a\times6 = 6a\). Also, the sum of the middle terms: \(6a+2a = 8a\), which matches the coefficient of \(a\) in the trinomial. So the missing number \(x\) is \(6\).

Answer:

6 (corresponding to the option "6")