How do polynomial identities apply to complex numbers?

A simple polynomial identity is:

x2−y2=(x−y)(x+y)

In fact this is just:

x2–1=(x−1)(x+1)

in disguise.

The benefit of such an identity is that it enables us to quickly find solutions when we equate the identity to  0 . In my example above this benefit is not easy to spot, we could use either half of the given identities and set it to  0 . But this is not always the case. For instance:

x2–2xy+y2=0

is already more difficult to solve than:

(x−y)(x−y)=0

Splitting a polynomial into factors, like in the examples above will give us an identity. The amount of factors we may find depend on the type of numbers we allow  x,y,⋯  to represent.

For instance you might have a hard time splitting:

x2+1

if we use  x∈R .

But if we allow  x∈C  we can factor this polynomial!

x2+1=(x+i)(x−i)

This also means that  x2+1=0  has two solutions:  x=±i . You may check this identity by using  i2=−1 .

In fact, a polynomial in one variable can always be split into polynomial factors of degree  1 . This is called the fundamental theorem of algebra. The extension of  R  to  C  enables us to factor a polynomial in one variable of degree  n  into exactly  n  factors, or which is the same thing, enables us to express all roots in the ‘field’ we use.

This difference, between being able to split a polynomial completely or only partly, depending on what values we allow for  x,y,⋯  does exist in a broader context. You will meet extensions, like from  R→C , in many varieties and truly appreciate what the word algebra means.

Step-by-step explanation:

answer: sine

step-by-step explanation: