Polynomial Identities

Identities

The identity $\left(a+b\right)^2=a^2+2ab+b^2$(a+b)2=a2+2ab+b2 and $\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3$(a+b)3=a3+3a2b+3ab2+b3 can be explained by the two diagrams below.  

 

The diagram on the left shows graphically what $\left(a+b\right)^2$(a+b)2 is. The identity can be written $\left(a+b\right)\left(a+b\right)$(a+b)(a+b), and what emerges is four partitioned areas. That is to say, two things can combine with two other things in four ways.

We combine these four things as products -  $a$a with $a$a, $a$a with $b$b, $b$b with $a$a and $b$b with $b$b. So the total area becomes $a^2+ab+ba+b^2$a2+ab+ba+b2, and because we know that $ab=ba$ab=ba, then the identity becomes $\left(a+b\right)^2=a^2+2ab+b^2$(a+b)2=a2+2ab+b2.

The diagram on the right shows $\left(a+b\right)^3$(a+b)3. This is two things combining with two things combining with two things and $2\times2\times2=8$2×2×2=8 things to think about.

The identity can be written $\left(a+b\right)\left(a+b\right)\left(a+b\right)$(a+b)(a+b)(a+b), and that is why we see the the task as determining the total volume of a 3 - dimensional cube of side $\left(a+b\right)$(a+b). 

There are in fact a total of $8$8 rectangular prism partitions. There are the cubes $a^3$a3 and $b^3$b3 in the top left and bottom right corners. There are also $3$3 prisms of volume $ab^2$ab2 and $3$3 prisms of volume $a^2b$a2b. The total volume is given by  $\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3$(a+b)3=a3+3a2b+3ab2+b3. 

The identity $\left(a+b\right)^4$(a+b)4 could be thought of as $4$4-dimensional, but of course we can't draw it. But what we do know is that there must be $2\times2\times2\times2=16$2×2×2×2=16 parts to consider, not all of which are different.

A pattern emerges with the quantities of each part that make up the total of $16$16 parts. The pattern was first discovered, at least in the western world by Blaise Pascal in the early 17th century. Pascal was able to provide the quantities for not only $\left(a+b\right)^2,\left(a+b\right)^3,\left(a+b\right)^4$(a+b)2,(a+b)3,(a+b)4, but for any positive integer power of $\left(a+b\right)$(a+b). Today it is known as Pascal's triangle.  

From that triangle we can write down the next two expansions:

$\left(a+b\right)^4$(a+b)4	$=$=	$a^4+4a^3b+6a^2b^2+4ab^3+b^4$a4+4a3b+6a2b2+4ab3+b4
$\left(a+b\right)^5$(a+b)5	$=$=	$a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5$a5+5a4b+10a3b2+10a2b3+5ab4+b5

Note that the total number of parts are $16$16 and $32$32 respectively.

Sum and difference of powers. 

A number of identities arise involving the sums and differences of two distinct powers.

For example the difference of the two squares $m^2$m2 and $n^2$n2, given by $m^2-n^2$m2−n2 gives rise to the identity $m^2-n^2=\left(m-n\right)\left(m+n\right)$m2−n2=(m−n)(m+n). 

As it turns out, this particular identity becomes extremely useful in areas of mathematics where simplification and factorisation is required. It is easily proven using a geometric diagram such as the one shown here:

In the diagram, a square of side $n$n has been cut out of a square of size $m$m. What remains can be divided into two rectangles. If the lower smaller rectangle is rotated and re-attached to the left hand side, we immediately see that the sum of the rectangles is given by $\left(m-n\right)\left(m+n\right)$(m−n)(m+n).

As an example of using this identity, consider this:

$87^2-13^2=\left(87-13\right)\left(87+13\right)=74\times100=7400$872−132=(87−13)(87+13)=74×100=7400

Also an expression like $\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)$(√7−2)(√7+2) can be simplified to $7-4=3$7−4=3. These are just two of the many uses for this identity.

But there are so many others, and they are all designed to help simplify and factorise more complicated mathematical expressions. We list just a few here:

$m^2+n^2=\left(m+n\right)^2-2mn$m2+n2=(m+n)2−2mn

$m^3-n^3=\left(m-n\right)\left(m^2+mn+n^2\right)$m3−n3=(m−n)(m2+mn+n2)

$m^3+n^3=\left(m-n\right)\left(m^2-mn+n^2\right)$m3+n3=(m−n)(m2−mn+n2)

     

Worked Examples

Question 1

Question 2

Question 3