Grade 8 Binomial Expansions-Reflection.

Name: Hassan Al-Thani

Class:8D

Date: November 3rd 2010

Teacher: Mrs.Tahr

Questions We know have a calculators that perform complex numerical calculations

easily, but in the ‘bad old days' it was all done with pencil paper.

Level 1-2:If you were an engineer 100 years ago, explain how our method may have been useful rather than just using long multiplication?

Level 3-4:At what point would our method be big and cumbersome? (ie. How many decimal places or what sorts of numbers would make us think twice about this method?)

Level 5-6:Can you give us some detailed explanations and examples of where long multiplication is more efficient than our expansion method?

Level 1-2 If I was an engineer 100 years ago, I would use our method it is more

useful instead of using long multiplication and the reason is that our method is easier shorter and they are not many decimal points. It is more useful because what we did is we converted (0.99) to the number (1-0.01) and to find (0.99)2 we must find (1-0.001)2 first. In order to find (1-0.001)2 we must use the rule (a-b)2=a2-2ab+b2 because it is easier then long multiplication when we use this rule (a-b)2=a2-2ab+b2 we will have 12=1, 2x1x0.01=0.02 and 0.012=0.0001 but if we use the long multiplication method it is a long way and more difficult because we will calculate bigger numbers and we will deal with much more decimal points such as (0.99)2=0.09x0.09+0.9x0.99.

After I have shown and compared both methods my results tell me that our method is easier, less decimal points to deal with but the long multiplication way is more difficult and you have to deal with much more decimal points

Level 3-4We will think twice about using this method, when

the numbers are less then 0.90 such as the number (0.89)2=(1-0.11)2 because the number (0.11)2=0.121 this number will be hard and cumbersome to find the sum because it has many decimal points, (1-0.11)2 =12-2x1x0.11+0.121. This is also suitable for the numbers greater than (1.10)2 such as the number (1.11)2=(1+0.11)2.

The suitable number for this rule are 0.90<x< 1.10So as you can see that the numbers less than 0.89

are cumbersome and the numbers greater than 1.11 are cumbersome also.

Level 5-6 The long multiplications is more efficient than our expansion method

when the numbers are less than 0.90 and greater than 1.10 for example 1.15, 1.19, 0.89, and 0.78.

(0.87)2=(0.87)x(0.87)= 0.87 (0.65)2=(0.65)x(0.65)=0.65 x0.87 0.65 0.0609 0.0325 0.6960 0.390 0.7569 0.4225As you can see the long multiplication is more efficient in some certain

numbers that I have mentioned above. The expansion method is very difficult to use with this number (0.87)2=(1-0.13)2=12-2x1x0.13+0.132 is very difficult to calculate.

Conclusion

This information says that sometimes our method will be really useful but not in all question, because in some situations our method will become cumbersome, but it will be more useful to use long multiplication and some situations or questions multiplication becomes cumbersome and for our method it is not.

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