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

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

2 Answers
Apr 14, 2018

Answer:

#(2x-1)^2#

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#

Apr 14, 2018

Answer:

#4x^2-4x+1 = (2x-1)^2# is a perfect square trinomial...

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 #4x^2 = (2x)^2# and #1 = 1^2# are both perfect squares. So we might think of putting #A=2x# and #B=1#. That would allow us to find:

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

That gives us #+4x# instead of the #-4x# that we want.

Note however that if we put #B=-1# instead of #B=1# then we still have #B^2 = (-1)(-1) = 1# as we need, but the middle term becomes #AB=(4x)(-1) = -4x# as we also want.

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 #A^2+2AB+B^2# or #A^2-2AB+B^2#, then you can recognise it as #(A+B)^2# or #(A-B)^2#.