91Ó°ÊÓ

The given trinomial is: $k^2−16k+64$

The First, middle, and last terms of the given trinomial are$k^2,-16k$ and 64 respectively.

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

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

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

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

$64={\left(8\right)}^2$

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

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

$−16k=−2\left(k\right)\left(8\right)$

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:

$$\begin{gathered} k^2−16k+64={\left(k\right)}^2−2\left(k\right)\left(8\right)+{\left(8\right)}^2 \\ =\left(k−8\right)\left(k−8\right)\left(âˆ\mu a^2−2ab+b^2=\left(a−b\right)\left(a−b\right)={\left(a−b\right)}^2\right) \\ ={\left(k−8\right)}^2 \end{gathered}$$

Therefore, the factorization of the given trinomial is ${\left(k−8\right)}^2$.