How do you factorise 4x^2-y^2?

4x² - y²

= (2x)² - y²

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

Hello Carys, recall a^2 - b^2 = (a+b)(a-b)
So (2x)^2 - y^2 = (2x+y)(2x-y) as a = 2x and b = y

How do you factorise 4x^2-y^2?

Note that this is a special pattern: a difference of squares (two perfect squares subtracted). A difference of squares is a special product in the form of (a^2 - b^2) and is factored into the form (a - b)(a + b), where here a = 2x and b = y (you can tell by taking the square root of both those terms)...so...


4x^2 - y^2

= (2^2)x^2 - y^2 <-- by exponent property: (ab)^n = a^n b^n..

= (2x)^2 - (y)^2 <-- seems familiar? That's because a = 2x and b = y...

= (2x - y)(2x + y) <-- answer...

4x^2 -y^2
=(2x)^2-y^2
=(2x-y)(2x+y)

(2x-y)(2x+y)

You don't "factorise" anything. There is no such thing as factorisation.
You factor. Now 4 x^2 - y^2 = (2x)^2 - y^2. Do you recognise that this is
a difference of squares of the form a^2 - b^2 = (a - b)(a + b), where a = 2x and b = y? Obviously not before posting your question. Otherwise you would have been able to simply write 4x^2 - y^2 = (2x - y)(2x + y)
and call it "End of story".

4x^2 – y^2 = 0
4x^2 = (2x)^2
(2x)^2 – y^2 = 0
Apply the algebraic property: a^2 – b^2 = (a+b) (a-b)
(2x + y) (2x-y)

Use the perfect square formula: a^2-b^2=(a+b)x(a-b)

So 4x^2-y^2=(2x+y)x(2x-y)

The x in between the two parentheses indicating multiplication, not another variable. Hope this helped.

4x^2-2xy+2xy-y^2

2x(2x-y)+y(2x-y)

(2x+y)(2x-y)

first of all, its FACTOR.

you should recognize that because it doesn't have three terms, only two in which both are squared, it will be a "difference of squares" problem.

when you have a difference of squares problem, its easy memorization.

take the square root of each term, the square root of the first term (4x^2) is 2x
the square root of the second term(y^2) is y.
Then you just put it in this form (A+B)(A-B) where A represents the square root of the first term and B represents the square root of the second term.

answer is: (2x+y)(2x-y) also you can always check your answer by FOIL-ing it out (multiplying it out) and seeing if you get the original problem.

[ 2x- y ] [ 2x + y ]