91Ó°ÊÓ

The given trinomial is: $4x^2+28x+49$

The First, middle and last terms of the given trinomial are$4x^2,28x$ and 49 respectively.

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

$4x^2={\left(2x\right)}^2$

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

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

$49={\left(7\right)}^2$

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

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

$28x=2\left(2x\right)\left(7\right)$

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:

$$\begin{gathered} 4x^2+28x+49={\left(2x\right)}^2+2\left(2x\right)\left(7\right)+{\left(7\right)}^2 \\ =\left(2x+7\right)\left(2x+7\right)\left(âˆ\mu a^2+2ab+b^2=\left(a+b\right)\left(a+b\right)={\left(a+b\right)}^2\right) \\ ={\left(2x+7\right)}^2 \end{gathered}$$

Therefore, the factorization of the given trinomial is ${\left(2x+7\right)}^2$.