91Ó°ÊÓ

The given trinomial is: $a^2+12a+36$

The First, middle and last terms of the given trinomial are $a^2,12a$and 36 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:

$36={\left(6\right)}^2$

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

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

$12a=2\left(a\right)\left(6\right)$

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 the 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="1647690661055" $$\begin{gathered} a^2+12a+36={\left(a\right)}^2+2\left(a\right)\left(6\right)+{\left(6\right)}^2 \\ =\left(a+6\right)\left(a+6\right)\left(âˆ\mu a^2+2ab+b^2=\left(a+b\right)\left(a+b\right)={\left(a+b\right)}^2\right) \\ ={\left(a+6\right)}^2 \end{gathered}$$Therefore, the factorization of the given trinomial is ${\left(a+6\right)}^2$.