How do you factor #121-22u+u^2# using the perfect squares formula?

Question

1611 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-27T05:56:03

Given -

#121-22u+u^2#

#u^2-22u+121=0#

Take the constant term to the right

#u^2-22u=-121#

Take half of the coefficient of #u#. Square it. Add the same to both sides.

#u^2-22u+121=-121+121#

#(u-11)^2=0#

#(u-11)(u-11)#

Answer 2 · 2017-02-27T06:05:13

#121-22u+u^2=(11-u)^2#

As the middle term is negative, let us compare

#121-22u+u^2# to the Left Hand Side of perfect square formula

#a^2-2ab+b^2=(a-b)^2#

It is observed that #color(red)121# is square #11^2# just like #color(red)(a^2)#.

Question is can we have #11# as #a#? .................(1)

we also have last term #color(blue)(u^2)# a square just like #color(blue)(b^2)#

Question is can we have #u# as #b#? .................(2)

The decision whether we can have (1) and (2)

is based on third term i.e. whether #-22u# is our #-2ab#.

We can check it. As #a# is #11# and #b# is #u#,

#-2ab=-2xx11xxu=-22u# and we can say that #-22u# is our #-2ab#.

and hence #121-22u+u^2=(11-u)^2#

Once our understanding is clear we can do the entire in short as the following

#color(red)121-22u+color(blue)(u^2)#

= #color(red)((11)^2)-2xxcolor(red)11xxcolor(blue)u+color(blue)(u^2)#

= #(color(red)11-color(blue)u)^2#