Difficult maths factorisation question?!

Factorise.
4x² - 81.

81ab² - 4ac²

Please explain if you could , how isit done and the steps.

They both involve difference of squares. You just need to know the form:
(f+d)(f-d) = f^2 - d^2
Test it out with FOIL. It works.

Now take a look at what you have for #1:
f^2 = 4x^2
d^2 = 81
I think you can figure out that f = 2x and d = 9.
So that leaves you with (2x+9)(2x-9)

#2 is slightly more tricky. First notice that both terms have a in them. So you can pull that out using distributive property:
a(81b^2 - 4c^2)
Focus on the part in the bracket:
f^2 = 81b^2
d^2 = 4c^2
f = 9b and d = 2c
That gives (9b+2c)(9b-2c)
Now add the a back in:
a(9b+2c)(9b-2c) is your answer.

4x² - 81= (2x+9)(2x-9) square of subtractions

81ab² - 4ac²=a(9b+2c)(9b-2c) take a common and square of subtractions

(2x-9)(2x+9) differance of squares

take out common factor
a(81b2-4c2)
(a)(9b+2c)(9b-2c) differance of squares

4x² - 81.
= (2x)² – (9)² ----- a² – b² = (a + b)(a – b)
= (2x + 9)(2x – 9)

81ab² - 4ac²
= a(81b² – c²)
= a[(9b)² – (c)²]
= a(9b + c)(9b – c)
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