Which is the complete factorization of x² + 14x + 49?

The trinomial $x^2+14x+49$ has $x^2$ and $7^2=49$ as its first and last terms, and the middle term $14x=2\cdot7\cdot x$ is twice the product of $x$ and $7$. So it is a perfect square trinomial: $(x+7)^2$, choice J. Choice F, $(x+7)(x-7)$, expands to $x^2-49$, a difference of squares. Choice G, $(x+14)(x+1)$, expands to $x^2+15x+14$. Choice H, $(x+49)(x+1)$, expands to $x^2+50x+49$. Only $(x+7)^2$ gives the middle term of $14x$ and the constant of $49$.