Question

which terms complete the factorization of $x^2 + 27x + 162$ represented by the model? \
\bigcirc 27, 9x, 18x \
\bigcirc 9, 9x, 18x \
\bigcirc 27, 9x, 27x \
\bigcirc 9, 9x, 27x

Explanation:

Step1: Analyze the area model

In the area model for factoring \(x^{2}+27x + 162\), we know that the product of the terms in the first row and first column is \(x^{2}\) (from \(x\times x\)). The last term \(162\) is the product of \(18\) and another term. Let's find that term: \(\frac{162}{18}=9\). So one of the missing terms (top - right cell) should be \(9\).

Step2: Find the other terms

The middle row (the row with \(x\)): the middle term of the quadratic is \(27x\), which is the sum of two terms. One term is from \(x\times9 = 9x\) and the other is from \(x\times18=18x\) (since \(9 + 18=27\)). Also, the cell below \(x^{2}\) (in the column with \(18\)) is \(18\times x=18x\)? Wait, no, the cell in the second row (with \(18\)) and second column: \(18\times x = 18x\)? Wait, no, the cell in the third row (with \(18\)) and second column: \(18\times9=162\)? Wait, no, let's re - express the quadratic. We know that \(x^{2}+27x + 162=(x + 9)(x + 18)\) (since \(9\times18 = 162\) and \(9 + 18=27\)). In the area model:

• The top - right cell (first row, third column) is \(9\) (because \(x\times9\) is part of the expansion, and also \(18\times9 = 162\)).
• The cell in the first row, second column (next to \(x\)) is \(9\) (wait, no, the first row: first cell \(x\), second cell? Wait, the columns: first column \(x\), second column? Wait, the rows: first row \(x\) (first cell), then the second cell in first row? Wait, maybe a better way: the area model is a 3x3 grid? Wait, no, it's a grid where the first row has \(x\) and another term, the second row has \(x\) (with \(x^{2}\) below it), and the third row has \(18\) and \(162\). Wait, actually, the correct way is:

The quadratic \(x^{2}+27x + 162\) can be factored as \((x + 9)(x + 18)=x^{2}+18x+9x + 162=x^{2}+27x + 162\). In the area model:

• The top - right cell (the cell in the first row, third column) is \(9\) (because when we multiply \(x\) (first row, first cell) and \(9\) (first row, third cell), and \(18\) (third row, first cell) and \(9\) (first row, third cell) gives \(162\)).
• The cell in the second row, third column (next to \(x^{2}\)) is \(18x\) (from \(x\times18\)).
• The cell in the first row, second column (next to \(x\)) is \(9\) (from the factor \(x + 9\), the second term in the first factor is \(9\)).
• The cell in the third row, second column (between \(18\) and \(162\)) is \(9x\) (from \(18\times x\)? No, from \(x\times9\)? Wait, no, when we expand \((x + 9)(x + 18)\), the terms are:
• \(x\times x=x^{2}\) (second row, second column)
• \(x\times9 = 9x\) (first row, second column? No, maybe the grid is:

Rows:

• Row 1: \(x\), \(9\)
• Row 2: \(x\), \(x^{2}\), \(9x\)
• Row 3: \(18\), \(18x\), \(162\)

Wait, no, the sum of the middle terms: \(9x+18x = 27x\). And the top - right term is \(9\), the other terms are \(9x\) and \(18x\). So the terms that complete the factorization are \(9\), \(9x\), \(18x\).

Answer:

B. 9, 9x, 18x