Is ${a^2} + 14a + 49$ a perfect square trinomial, and how do you factor it?

Hint: As we know that an expression which can be obtained from the square of the binomial equation is called a perfect square trinomial. An expression is said to be a perfect square trinomial if it can take the form of $a{x^2} + bx + c$ and also satisfies the discriminant of the quadratic formula i.e. ${b^2} = 4ac$. The formula square formula can be verified if it takes the form of an identity or we can say that an identity satisfies it like ${(a + b)^2} = {a^2} + 2ab + {b^2}$.

Complete step-by-step solution:
As per the given question we have an equation ${a^2} + 14a + 49$, we are asked to check whether it is a perfect square trinomial or not.
Now the given quadratic polynomial can be the only perfect square trinomial if it can be written in the form ${(a + b)^2}$. Now by considering the coefficient of ${a^2}$ and $a$, we have constant terms. We can write $a = 1,b = 14,c = 49$. We will now apply the formula of discriminant we have, $ \Rightarrow {D^2} = {b^2} - 4ac$, By substituting the values we have $ \Rightarrow D = {(14)^2} - 4 \times 1 \times 49$
$D = 196 - 196 \Rightarrow D = 0$. Therefore the value of discriminant is zero. We can say that the above equation is a perfect square trinomial. Now we will factor the given quadratic polynomial. We can write the given expression as: ${a^2} + 14a + 49 = {(a)^2} + 2 \times a \times 7 + {(7)^2}$.
We can see that the above expression is of the form ${(a + b)^2}$ where $a = a,b = 7$.
So it can be written as ${(a + 7)^2}$.

Hence it is a perfect square trinomial and ${(a + 7)^2}$ is the factored form of the given quadratic expression.

Note: We should note that the trinomial in the expression means a three term polynomial not a cubed polynomial. A perfect square binomial is a trinomial which when factored gives us the square of a binomial. Whenever we take a binomial and multiply it to itself, it gives us a perfect square trinomial.