Need help factoring a^4 - 625?

Actually, an equation can be many things.

This is a quartic (4th power).

However, it is also a difference of squares:
a^4 is the square of a^2
625 is the square of 25

Model for difference of squares:
x^2 - y^2 = (x+y)(x-y)

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

Look! the second factor is also a difference of squares!
I'll let you do it.

[1]
a^4-625
=(a^2)^2-(25)^2
=(a^2+25)(a^2-25)
(a^2+25){(a)^2-(5)^2}
=(a^2+25)(a+5)(a-5)

Because both of the terms are perfect squares, this is a difference of two squares. It is of the form A^2 - B^2.

For a variable, you know it's a perfect square if it's exponent is an even number. In this case, the variable 'a' has an exponent of 4, so it's a perfect square. The square root is obtained by dividing the exponent by 2, so the square root of a^4 is a^2.

For a constant, you can always check by using a calculator if the square root of a number is an integer. The square root of 625 is 25.

The factors of a polynomial of the form A^2 - B^2 are (A-B) and (A+B).

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

Now a^2-25 is another perfect square. So, it can still be factored into (a-5)(a+5). So

a^4 - 625
=(a-5)(a+5)(a^2+25)

This is the completely factored form.

a^4 - 625
= (a^2 + 25)(a^2 - 25)
= (a^2 + 25)(a + 5)(a - 5)

(a² - 25)(a² + 25)
(a - 5)(a + 5)(a² + 25)

a^4 - 625
rule is x² – y² = (x + y)(x – y)
a^4 - 625
= (a²)² – (25)²
= (a² + 25)(a² – 25)
= (a² + 25)(a + 5)(a – 5)