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

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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

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

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

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

How do you factor 121-22u+u^2121-22u+u^2 using the perfect squares formula?