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

Feb 27, 2017

Given -

$121 - 22 u + {u}^{2}$

${u}^{2} - 22 u + 121 = 0$

Take the constant term to the right

${u}^{2} - 22 u = - 121$

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

${u}^{2} - 22 u + 121 = - 121 + 121$

${\left(u - 11\right)}^{2} = 0$

$\left(u - 11\right) \left(u - 11\right)$

Feb 27, 2017

$121 - 22 u + {u}^{2} = {\left(11 - u\right)}^{2}$

#### Explanation:

As the middle term is negative, let us compare

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

${a}^{2} - 2 a b + {b}^{2} = {\left(a - b\right)}^{2}$

It is observed that $\textcolor{red}{121}$ is square ${11}^{2}$ just like $\textcolor{red}{{a}^{2}}$.

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

we also have last term $\textcolor{b l u e}{{u}^{2}}$ a square just like $\textcolor{b l u e}{{b}^{2}}$

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 $- 22 u$ is our $- 2 a b$.

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

$- 2 a b = - 2 \times 11 \times u = - 22 u$ and we can say that $- 22 u$ is our $- 2 a b$.

and hence $121 - 22 u + {u}^{2} = {\left(11 - u\right)}^{2}$

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

$\textcolor{red}{121} - 22 u + \textcolor{b l u e}{{u}^{2}}$

= $\textcolor{red}{{\left(11\right)}^{2}} - 2 \times \textcolor{red}{11} \times \textcolor{b l u e}{u} + \textcolor{b l u e}{{u}^{2}}$

= ${\left(\textcolor{red}{11} - \textcolor{b l u e}{u}\right)}^{2}$