The **difference** of two squares occurs when the middle term or "b" of your polynomial is absent.

Looking at the standard form of

ax^{2}+bx+c = 0

The "b" term is absent when the sum of the cross multiplication is zero:

b = (a*c) + (a*-c) = 0

so the the "c" term would have to be the product of a single number, one positive and one negative.

In other words, and using a = 1 for simplicity

x^{2}-c^{2} = 0

to show you the series using small integers:

if c =1 then x^{2}-1 = 0 factors to (x**+**1)(x**-**1) = 0

if c =2 then x^{2}-4 = 0 factors to (x**+**2)(x**-**2) = 0

if c = 3 then x^{2} - 9 = 0 factors to (x**+**3)(x**-**3) = 0 and so on.

Notice how the signs alternate so that you are squaring the number by its positive and negative value. That means that when you multiply it out the cross product sum or middle term "b" has equal and opposite values and will then drop out of the expression.

(x+c)(x-c) = x^{2} + cx - cx - c^{2} = x^{2} - c^{2}

The **sum** of two squares involves complex conjugates using the imaginary number or operator "i"

"i" is used to describe a number that satisfies the equation x^{2} = -1 so that x = √-1 = i

Every complex number has a corresponding conjugate and exist in pairs, this is typically written as

(a ± bi) but to use the notation from the example above i'll stick with x and c

**(x+ci)(x-ci) = 0**

multiplying that out yields

x^{2} + cxi - cxi - c^{2}i^{2} = 0

Now knowing that i^{2} = -1 you get the answer you are looking for

x^{2} + c^{2} = 0

So for a positive integer series of c

where c = 1 then x^{2} + 1 = 0 factors to (x**+**i)(x**-**i)

if c =2 then x^{2} + 4 = 0 factors out to (x**+**2i)(x**-**2i)

if c = 3 then x^{2} + 9 = 0 factors out to (x**+**3i)(x**-**3i)

Again notice how the signs alternate so that you have one positive value and one negative value.

Michael B.

Also, in my example above for the difference of squares I was using "perfect" squares. If you look at the equation itself

x

^{2}- c^{2}= 0 and then add c^{2}to both sides you getx

^{2}= c^{2}and that x = ±vc^{2}so for something like x

^{2}- 10 = 0 you would get x = ±v10.In factor form:

(x+v10)(x-v10) = 0

02/11/13