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

Question

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

Abigail Allen · Accepted Answer

#(2x-1)^2# #"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#

Savannah Anderson · Answer

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

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#.

Nathan Axel · Answer

undefined To factorize $4x^2 - 4x + 1$, recognize that it's a perfect square trinomial. Its factored form is $(2x - 1)^2$.