Factorise as far as possible into Prime Factors:
5x^8 - 80y^4
first factor out a 5
5(x^8 - 16y^4)
now you should see a difference of squares
5(x^4 - 4y^2)(x^4 + 4y^2)
and once more
5(x^2 - 2y)(x^2 + 2y)(x^4 + 4y^4)
I believe Reiny meant to write
5(x^2 - 2y)(x^2 + 2y)(x^4 + 4y^2)
for the last step. The last y exponent is 2.
thanks drwls for catching that.
Time to get new reading glasses at the "Dollar Store", lol
To factorize the given expression, 5x^8 - 80y^4, into prime factors, we can use the difference of squares formula:
a^2 - b^2 = (a + b)(a - b)
In this case, we can rewrite the expression as:
(5x^8)^2 - (8y^4)^2
Now, let's identify a and b in the difference of squares formula:
a = 5x^8
b = 8y^4
Now, we can substitute these values into the formula:
(5x^8 + 8y^4)(5x^8 - 8y^4)
Now, let's focus on factoring the first parentheses, (5x^8 + 8y^4), further. Since 5x^8 and 8y^4 do not share any common factors, it cannot be factored anymore.
However, we can factor the second parentheses, (5x^8 - 8y^4), further using the difference of cubes formula.
The difference of cubes formula is given by:
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
In this case, we can rewrite the expression (5x^8 - 8y^4) as:
(5x^2)^3 - (2y^2)^3
Now, let's identify a and b in the difference of cubes formula:
a = 5x^2
b = 2y^2
Now, we can substitute these values into the formula:
(5x^2 - 2y^2)((5x^2)^2 + (5x^2)(2y^2) + (2y^2)^2)
The first parentheses, (5x^2 - 2y^2), cannot be factored further because the terms do not have any common factors.
The second parentheses, ((5x^2)^2 + (5x^2)(2y^2) + (2y^2)^2), can be simplified as follows:
(25x^4 + 10x^2y^2 + 4y^4)
So, the prime factorization of 5x^8 - 80y^4 is:
(5x^2 - 2y^2)(25x^4 + 10x^2y^2 + 4y^4)