# Homework Help: Factoring Involving the Difference of Two Squares

1. Sep 18, 2017

### DS2C

1. The problem statement, all variables and given/known data
Completely factor:

$$5abc^4-80ab$$

2. Relevant equations

N/A

3. The attempt at a solution

$$5abc^4-80ab$$
$$5ab\left(c^4-16\right)$$
$$5ab\left(c^2+4\right)\left(c^2-4\right)$$

The correct solution is $5ab\left(c^2+4\right)\left(c+2\right)\left(c-2\right)$
I can see why it factored out the $\left(c^2-4\right)$ further, but why isn't it required to do so with the $\left(c^2+4\right)$ as well?

2. Sep 18, 2017

### Staff: Mentor

Because $c^2 - 4$ is a difference of squares while $c^2 + 4$ is the sum of two squares. The latter can't be factored into two linear binomials whose coefficients are real. It can, however, be factored into complex numbers, as $(c + 2i)(c - 2i)$, but in ordinary factoring, you're only concerned with factors with real coefficients.

3. Sep 18, 2017

### DS2C

Thanks for that quick response, Mark. But why can the difference between two squares be factored with real numbers however the sum between two squares cannot?

4. Sep 20, 2017

### Staff: Mentor

Let's look at the equation $x^2 - 1 = 0$. I don't think you have any problem with recognizing that the equation could be rewritten as (x - 1)(x + 1) = 0.

Now consider the equation $x^2 + 1 = 0$. Is is possible to rewrite this as (x + a)(x + b) = 0, assuming that both a and b are real numbers? If so, we must have a*b = 1 and a + b = 0 (the latter since the coefficient of x is 0).

If a + b = 0, then a and b must be opposites (i.e., additive inverses), which would also include the possibility that a = b = 0.
If ab = 1, then a and b need to be reciprocals (multiplicative inverses) of one another. This rules out the possibility that either one could be zero. It also says that both a and b have to be positive, or both have to be negative. Each of these latter conclusions rules out the possibility of a and b being opposites.

The upshot is that there are no real numbers a and b so that (x + a)(x + b) = 0. If I relax the condition that a and b must be real, it turns out that there are solutions for a and b; namely, a = i and b = -i, where i is the imaginary number $\sqrt{-1}$. Thus, allowing coefficients from the complex numbers, $x^2 + 1$ can be factored into $(x + i)(x - i)$.

5. Sep 24, 2017

### DS2C

Thanks a lot for that reply Mark. Been real busy with schoolwork so I got to this a little late, but it makes sense. I'll keep this in mind when I eventually get to imaginary numbers.

6. Sep 24, 2017

### DS2C

I did a little fiddling and multiplied out $\left(c+2\right)\left(c-2\right)$ and came up with $\left(c^2-4\right)$.
In multiplying $\left(c+2\right)\left(c+2\right)$, I got $c^2+4c+4$.
Is this another way of saying it as well?

7. Sep 24, 2017

### Staff: Mentor

Well, sort of. The multiplication you're doing is the opposite operation of factoring. You found that (c + 2)(c - 2) = c2 - 4, but there aren't any real numbers a and b for which (x - a)(x - b) = x2 + k2.