The
method of detached coefficients is simply a way of dealing with
algebra by dropping the algebraic variables and simply using the
numbers, keeping them in their correct places. Some examples follow.

Binomial Coefficients with a Calculator

The
Method of Detached Coefficients involves detaching the coefficients of
polynomial and dealing with the coefficients only. For instance: (1+x)^{2}=1+2x+x^{2} 11^{2}=121

(1+x)^{3}=1+3x+3x^{2}+x^{3} 11^{3}=1331

(1+x)^{4}=1+4x+6x^{2}+4x^{3 }+x^{4} 11^{4}=14641

That
is, we can drop the x's, y's etc, and use the coefficients alone. In
the above cases, we find the coeffients simply by using 11 instead of
(1+x), and using arithmtic to expand our binomial (or whatever).

After 11^{4}, the calculator does not give us the clear result we seek:

11^{5}=161051

This reminds us that we need to keep our numbers separate: 101^{5}=10510100501

This
means we soon exceed the capacity of the ten-digit calculator, and need
to use Windows Calculator. However, adding a zero, keeps the
coefficients separate, so we can read of the results, and write: 101^{5}=10510100501, as (a+b)^{5}=a^{5}+5a^{4}b+10a^{3}b^{2}+10a^{2}b^{3}+5ab^{4}+b^{5}However, an ordinary calculator gives: 101^{5}=105101005 from which we can guess the last digit is 1.

The important thing is to keep the coefficients separate. A final example:

We cannot so easily use the calculator in the case of multinomial coefficients of the form:(a+b+c)^{2} (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2ac+2bc

We cannot do this because if we write it as 111, for instance, with a=100, b=10, and c=1, we get confused between b^{2}=100 and ac=100, and 111^{2}=12321 with the 3 being a confusion between b^{2} and 2ac.

To draw out the difference we can write: 10001001^{2}=100020021002001 where a=10,000,000, b=1000, and c=1.

The first 1 is a^{2}, the first 2 is 2ab, the next 2ac. The second 1 is b^{2} and the final 2 is 2bc, and the final 1 c^{2}. (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2ac+2bc

Another example: 10001001^{3}=1000300330061033003001

This implies that (a+b+c)^{3}=a^{3}+b^{3}+c^{3}+3a^{2}b+3a^{2}c+3ab^{2}+3ac^{2}+3b^{2}c+3bc^{2}+6abc

Multinomials of the form (1+x+x^{2})^{n}

Multinomials of the form (1+x+x^{2}) have a basic order, so: 111^{2}=12321

Ken's book is packed with examples and explanations that enable you to discover more than 150 techniques to speed up your arithmetic and increase your understanding of numbers. Paperback and Kindle: