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ms. wilson draws a model of the factorization of a polynomial with inte…

Question

ms. wilson draws a model of the factorization of a polynomial with integer factors. her model is partially complete. model grid with cells: n, (empty); (empty), n², (empty); 5, 5n, 40 which equation is represented by ms. wilson’s model? ○ n² + 3n + 40 = (n - 8)(n - 5) ○ n² + 13n + 40 = (n + 8)(n + 5) ○ n² + 40n + 13 = (n + 8)(n + 5) ○ n² + 40n + 3 = (n - 8)(n - 5)

Explanation:

Step1: Analyze the table

Looking at the table, the bottom row has 5, \(5n\), and 40. The middle column has \(n\), \(n^2\), \(5n\). So we can find the factors. The first column: 5 and the middle cell \(n^2\) related? Wait, the bottom row: 5 times something is 40? \(40\div5 = 8\), so the right bottom cell's factor with 5 is 8. And the top middle is \(n\), so the top right should be \(8n\) (since 5 times 8 is 40, and \(n\) times 8 is \(8n\)). Then the middle row: \(n^2\) is \(n\times n\), so the first middle cell is \(n\), so the first column middle is \(n\) (since \(n\times n = n^2\)). Now, the polynomial is the sum of all cells: \(n^2 + 8n + 5n + 40\)? Wait, no, the table is a area model, so the length and width. The bottom row: 5 and 8 (since 5*8=40), and the middle column: \(n\) and 5? Wait, no, the middle row has \(n^2\), which is \(n\times n\), so one side is \(n + 5\) (since bottom row first is 5) and the other side is \(n + 8\) (since bottom row third is 40, 40/5=8). So the polynomial is \((n + 8)(n + 5)=n^2 + 13n + 40\).

Step2: Check the options

Now check the options. The second option is \(n^2 + 13n + 40=(n + 8)(n + 5)\), which matches.

Answer:

B. \(n^2 + 13n + 40 = (n + 8)(n + 5)\) (assuming the options are labeled A, B, C, D with B being this one)