91Ó°ÊÓ

The given trinomial is: $a^2−22a+121$

The First, middle, and last terms of the given trinomial are$a^2,-22a$ and 121 respectively.

The first term of the given trinomial can be written as:

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

Therefore, the first term of the given trinomial is a perfect square.

The last term of the given trinomial can be written as:

$121={\left(11\right)}^2$

Therefore, the last term of the given trinomial is a perfect square.

The middle term of the given trinomial can be written as:

$−22a=−2\left(a\right)\left(11\right)$

Therefore, the middle term is twice the product of the square roots of the first term and last term.

As, the first and last terms of the given trinomial are a perfect square and the middle term is twice the product of the square roots of first term and last term.

Therefore, yes, the given trinomial is a perfect square trinomial.

It is known that:

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

It can be noticed that:

role="math" localid="1647692536357" $$\begin{gathered} a^2−22a+121={\left(a\right)}^2−2\left(a\right)\left(11\right)+{\left(11\right)}^2 \\ =\left(a−11\right)\left(a−11\right)\left(âˆ\mu a^2−2ab+b^2=\left(a−b\right)\left(a−b\right)={\left(a−b\right)}^2\right) \\ ={\left(a−11\right)}^2 \end{gathered}$$

Therefore, the factorization of the given trinomial is ${\left(a−11\right)}^2$.