Question

Ahmed found the product of (23,/g2−6n)2
using the area model. His partial work is shown. What is the partial product that goes in the orange box? What is the partial product that goes in the blue box? What is the product of (23g2−6n)2
written in standard form?(3 points)

Answer 1

To find the product of \((23g^2 - 6n)^2\) using the area model, we can use the formula for squaring a binomial, which is given by:

\[
(a - b)^2 = a^2 - 2ab + b^2
\]

Here, we can identify:
- \(a = 23g^2\)
- \(b = 6n\)

Applying the formula, we calculate each term:

1. **\(a^2\)**: \((23g^2)^2 = 529g^4\)
2. **\(b^2\)**: \((6n)^2 = 36n^2\)
3. **\(-2ab\)**: \(-2 \cdot (23g^2) \cdot (6n) = -276g^2n\)

Now we can summarize all the components:

- The term that goes in the **orange box** (which typically represents \(a^2\)) is **\(529g^4\)**.
- The term that goes in the **blue box** (which typically represents \(b^2\)) is **\(36n^2\)**.

Putting it all together, the product \((23g^2 - 6n)^2\) can be written in standard form as:

\[
529g^4 - 276g^2n + 36n^2
\]

Thus, the final answer for \((23g^2 - 6n)^2\) in standard form is:

\[
529g^4 - 276g^2n + 36n^2
\]