Asked by anonymous
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.
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.
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