# How to factorise #4x^2-4x+1#?

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Don't see how I can factorise to #(2x-1)^2# ? Thank you so much!

Don't see how I can factorise to

##### 2 Answers

#### Answer:

#### Explanation:

#"for a quadratic in "color(blue)"standard form"#

#•color(white)(x)ax^2+bx+c ;a!=0#

#"consider the factors of the product ac which sum to b"#

#4x^2-4x+1" is in standard form"#

#"with "a=4,b=-4" and "c=1#

#rArrac=4xx1=4" and - 2 , - 2 sum to - 4"#

#"split the middle term using these factors"#

#4x^2-2x-2x+1larrcolor(blue)"factorise in groups"#

#=color(red)(2x)(2x-1)color(red)(-1)(2x-1)#

#"take out the common factor "(2x-1)#

#=(2x-1)(color(red)(2x-1))#

#rArr4x^2-4x+1=(2x-1)^2#

#### Answer:

#### Explanation:

Given:

#4x^2-4x+1#

This is an example of a perfect square trinomial.

Let's take a look at what happens when you square a binomial using FOIL to help us:

#(A+B)^2 = (A+B)(A+B)#

#color(white)((A+B)^2) = overbrace(A * A)^"First"+overbrace(A * B)^"Outside"+overbrace(B * A)^"Inside"+overbrace(B * B)^"Last"#

#color(white)((A+B)^2) = A^2+AB+AB+B^2#

#color(white)((A+B)^2) = A^2+2AB+B^2#

In our example, note that

#(2x+1)^2 = 4x^2+4x+1#

That gives us

Note however that if we put

So:

#(2x-1)^2 = (2x)^2+2(2x)(-1)+(-1)^2 = 4x^2-4x+1#

More generally, we can write:

#(A-B)^2 = A^2-2AB+B^2#

So given any quadratic in standard form, if the first and last terms are perfect squares and it matches the pattern