# what exactly is a prime factor would x-5 x+5 and x2+25 all be prime factors of x4-625?

Yes, I believe those would be the prime factors.

x^4 - 625 can be factored into

(x^2 -25)(x^2 + 25)

and once again into

(x-5)(x+5)(x^2 + 25)

That is what you have done.

Any additional factoring would require imaginary numbers, using

x^2 + 25 = (x + 5i)(x - 5i)

where i is the square root of -1

I have never heard of the term "prime factor" with regard to polynomials, but it may refer to all the factors with real roots, which would be

(x-5)(x+5)(x^2 + 25)

in this case.

## A prime factor is a prime number that divides another number evenly without leaving a remainder. In the context of algebra, prime factors refer to irreducible factors of a polynomial that cannot be further factored into simpler polynomials.

To determine if x - 5, x + 5, and x^2 + 25 are prime factors of x^4 - 625, you can use the factoring method.

First, express x^4 - 625 as a difference of squares:

x^4 - 625 = (x^2)^2 - (25)^2 = (x^2 - 25)(x^2 + 25)

Next, you can factor x^2 - 25 as a difference of squares:

x^2 - 25 = (x - 5)(x + 5)

So far, we have factored x^4 - 625 as (x^2 - 25)(x^2 + 25), where x^2 - 25 further factors into (x - 5)(x + 5).

To determine if x - 5, x + 5, and x^2 + 25 are prime factors, you can check if they divide x^4 - 625 evenly. If they do, they are prime factors.

For x - 5 to be a prime factor, the polynomial (x^2 + 25) should be divisible by (x - 5). Similarly, for x + 5 to be a prime factor, (x^2 + 25) should be divisible by (x + 5).

If you perform polynomial division, you will find that (x^2 + 25) is divisible by (x - 5) and (x + 5) without leaving any remainder.

Therefore, x - 5, x + 5, and x^2 + 25 are indeed prime factors of x^4 - 625.

To summarize, the prime factors of x^4 - 625 are (x - 5), (x + 5), and (x^2 + 25).