How do you factor the trinomial \[4{x^2} + 20x + 25 = 0\]?

Hint:Factoring reduces the higher degree equation into its linear equation. In the above question, we need to reduce the trinomial into its simplest form in such a way that addition of products of the factors of first and last term should be equal to the middle term.

Complete step by step solution:
Trinomial is a polynomial consisting of three terms or monomials. The above equation is a trinomial since it has three terms.
\[a{x^2} + bx + c\] is a general way of writing quadratic equations where a, b and c are the numbers.
In the above expression,
a=4, b=20, c=25
\[4{x^2} + 20x + 25 = 0\]
First step is by multiplying the coefficient of \[4{x^2}\] and the constant term 25, we get $100{x^2}$.
After this, factors of $100{x^2}$ should be calculated in such a way that their addition should be equal to $20x$.
Factors of 100 can be 10 and 10.
where $10x + 10x = 20x$.
So, further we write the equation by equating it with zero and splitting the middle term according to the factors.
\[
\Rightarrow 4{x^2} + 20x + 25 = 0 \\
\Rightarrow 4{x^2} + 10x + 10x + 25 = 0 \\
\]
Now, by grouping the first two and last two terms we get common factors.
\[
\Rightarrow 2x\left( {2x + 5} \right) + 5\left( {2x + 5} \right) = 0 \\
\Rightarrow \left( {2x + 5} \right)(2x + 5) = 0 \\
\]
Taking 2x common from the first group and 5 common from the second we get the above equation.
We can further solve it and taking square root we get,
\[
{\left( {2x + 5} \right)^2} = 0 \\
\\
\]
Therefore, we get the above solution for the equation.

Note: In quadratic equation, an alternative way of finding the factors is by using the formula of sum of the terms a and b \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\].So by
converting the equation according to the formula we get \[{(2x)^2} + 2 \times 2x \times 5 + {(5)^2} = {(2x + 5)^2}\]