Factorise Fully?

Take out 2 as a common factor:
2(x²-25y²)
(x²-25y²) is the "difference of two squares" and you factorise it like this:
(x-5y)(x+5y)
your final answer therefore is:
2(x-5y)(x+5y)

2x^2 - 50y^2
= 2(x^2 - 25y^2)
= 2(x + 5y)(x - 5y)

this is a thing when you have the same variables you can subtract so the answer is i think 48

2(x - 5) (x + 5)

The above equates to 2q(p-q) which would be whether p and q are integers to initiate with. you may do (p-q) first, then multiply this by potential of q. Then it extremely is not correct whether q(p-q) is peculiar or perhaps by using fact any integer accelerated by potential of two is even. in case you have been questioning what might take place if p=q or p and q are 0, 0 is an integer and it extremely is even. Fretty, you have further q^2 on quite of taking it away.

2(x² - 25y²)
2(x - 5y)(x + 5y)

2x² - 50y² = 2(x² - 25y²)
= 2(x - 5y)(x + 5y)

Common factor of 2, then difference of two squares

= 2(x^2 - 25y^2)

= 2(x + 5y)(x - 5y)