Factorise: 4x^2 - 20x + 25?

Hi,

This is the difference of perfect squares, giving factors of (2x - 5)(2x - 5).

I hope that helps!! :-)

4x^2 - 20x + 25
= (2x - 5)(2x - 5)
= (2x - 5)^2

(2x - 5)(2x - 5)
(2x - 5) ²

i think the sum is wrong

4x^2 - 20x + 25
4x^2 -10x-10x+25
=2x (2x-5)-5(2x-5)
=(2x-5)^2

4x^2 - 20x + 25

= (2x - 5)(2x - 5) -----> FOIL METHOD

= (2x - 5)^2 -------> Factors

It is asked factorise (not asked only factors)
4x^2 – 20x + 25
product of coefficient of 1st and 3rd term = 4 × 25 = 100
Find the factors of 100 such that their sum or difference is coefficient of middle term.
100 = 10 × 10 and 10 + 10 = 20
But we want – 20 and (– 10 – 10 ) = – 20
4x^2 – 20x + 25
= 4x^2 – 10x – 10 x + 25
= 2 x (2x – 5) – 5 (2x – 5)
= (2x – 5)(2x – 5)

it's not the diference of two squares......a^2 + 2ab + b^2 = (a+b)^2 the diference between 2 squares is a^2 - b^2 = (a-b)(a+b) and yes....the answer is (2x-5)^2

The difference of two squares rule with the second term having a negative coefficient.
a^2 - 2ab + b^2 = (a-b)^2
in this case a = 2x ans c=5
(2x-5)^2

What's the big deal? it is (2x-5)^2