Answer:
Yes,a similar formula is used to find all the products
Explanation:
The formula applied in this case is;
(a+b)²= (a+b) (a+b)= a(a+b)+b(a+b) = a²+ab+ab+b²= a²+2ab+b²
In the first one;
![(x+3)^2= (x+3) (x+3) = x(x+3)+3(x+3) = x^2 +3x+3x+9 = x^2 +6x +9](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ipebq0hh035jeabjb3os8v061fser0tlon.png)
In the second one;
![(x+4)^2 = (x+4) (x+4) = x(x+4)+ 4(x+4) = x^2 +4x+4x+16 = x^2 +8x+16](https://img.qammunity.org/2020/formulas/mathematics/middle-school/7frg54ezabhyxzu864qtjxjml9wdfahz1u.png)
⇒This is the same for the third and fourth product.