how can you determine whether that polynomial is a difference of two squares?

Question

How do you factor a difference of two squares? Is there a similar strategy for
the sum of two squares?

Answer 1

Difference of two squares factors in this way: a² - b² = (a + b)(a -b)

You can determine if it's a difference of two squares, by seeing if both terms are perfect squares, and can be written in the form a² - b². Michael already showed how to factor it.

eg. 4x² - 25 can be written as (2x)² - 5² 

now notice we have it in that form a² - b² where a = 2x and b = 5.

(2x)² - 5² = (2x - 5)(2x+5)

 

Another example

9x² - 16

This can be written as (3x)² - 4²

Now try factoring it with the difference of two squares.

Answer 2

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

Looking at the standard form of

ax²+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

x²-c² = 0

to show you the series using small integers:

if c =1 then x²-1 = 0 factors to (x+1)(x-1) = 0

if c =2 then x²-4 = 0 factors to (x+2)(x-2) = 0

if c = 3 then x² - 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) = x² + cx - cx - c² = x² - c²

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 x² = -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

x² + cxi - cxi - c²i² = 0

Now knowing that i² = -1 you get the answer you are looking for

x² + c² = 0 

So for a positive integer series of c

where c = 1 then x² + 1 = 0 factors to (x+i)(x-i)

if c =2 then x² + 4 = 0 factors out to (x+2i)(x-2i)

if c = 3  then x² + 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.