Ahmed found the product of (23g2−6n)2

Question

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 start by recognizing it as a binomial squared. The general formula for squaring a binomial $(a-b{)}^2$ is $a^2-2ab+b^2$. In this case, we have $a=23g^2$ and $b=6n$.

1. Finding the partial product for the orange box (area for $a^2$): $$a^2=(23g^2{)}^2=529g^4$$

2. Finding the partial product for the blue box (area for $-2ab$): $$-2ab=-2(23g^2)(6n)=-276g^{2}n$$

3. Finding $b^2$ (which goes in the last part of the area model) for the complete product: $$b^2=(6n{)}^2=36n^2$$

Now, we can combine these results to write the complete expansion of $(23g^2-6n{)}^2$ in standard form: $$(23g^2-6n{)}^2=a^2-2ab+b^2=529g^4-276g^{2}n+36n^2$$

Thus, the complete standard form of the product is:

$$\boxed{{\displaystyle 529g^4-276g^{2}n+36n^2}}$$

In summary:

• The partial product in the orange box is $529g^4$.
• The partial product in the blue box is $-276g^{2}n$.
• The product in standard form is $529g^4-276g^{2}n+36n^2$.