How would you factorise a²-25?

(a - 5) (a + 5)

A^2-25 is simply the difference of two squares

the answer would be
(a+5) (a-5)

to confirm it we do
A x A= A^2
A x -5= -5A
A x +5= +5A
5 x -5= -25

which if u multiplied it all out it would be A^2 - 25

a²-25 = a²-5²
=> (a-5)(a+5)

a²-25
it is in form of a ^2 - b^ 2 = (a+b) (a-b)

so a²-25= a^2 - 5^2 which can be wriiten as (a+5) (a-5)

factor a²-25
= ( x+ 5 ) ( x- 5)

a^2 - b^2 = (a + b)(a - b)

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

This is what's known as a "difference of 2 squares."

The general formula to factorize these types of problems is as follows:
a² - b² = (a + b)(a - b)

In this case, b² = 25. By taking the root of that we see that b = 5.

So in factorizing you get:
a² - b² = (a + b)(a - b)
a² - 25 = (a + 5)(a - 5)

i believe its (a+5)(a-5) ....you would multiply the a's and then the fives, one positive one negative will give you a negative. If you need a positive 25 it would be two negative fives. make sense?

(a+5)(a-5)

A x A times -25