How do you factor $25{x^2} + 10x + 1 = 9$ ?

Hint: Write the given equation in the standard form $a{x^2} + bx + c = 0$ by transferring \[9\] to left-hand side. The resultant equation will be $25{x^2} + 10x - 8 = 0$.
Write the factors of the equation.
Write $10x = 20x - 10x$ because $20x \times ( - 10x) = - 200{x^2}$which is same as $25{x^2} \times ( - 8) = - 200{x^2}$.
Take the common term outside and equate each factor to $0$.

Complete step-by-step answer:
Consider the given equation is $25{x^2} + 10x + 1 = 9$.
Write the given equation in the standard form $a{x^2} + bx + c = 0$
where $a,b,$ and $c$ are real numbers.
Transfer $9$ to the left side of the equation.
$25{x^2} + 10x + 1 - 9 = 0$
$ \Rightarrow 25{x^2} + 10x - 8 = 0$
Write $10x = 20x - 10x$ because $20x \times ( - 10x) = - 200{x^2}$which is same as $25{x^2} \times ( - 8) = - 200{x^2}$.
$ \Rightarrow 25{x^2} + 20x - 10x - 8 = 0$
Take the common factors outside of the bracket.
$ \Rightarrow 5x(5x + 4) - 2(5x + 4) = 0$
$ \Rightarrow (5x - 2)(5x + 4) = 0$
Equate each factor to $0$.
$ \Rightarrow (5x - 2) = 0$ or $(5x + 4) = 0$

Final Answer: The factors of the equation $25{x^2} + 10x + 1 = 9$ are $(5x - 2)(5x + 4) = 0$.

Note:
The standard form of the quadratic equation is $a{x^2} + bx + c = 0$
where $a,b,$ and $c$ are real numbers.
The common mistake by the Students is that they write $25{x^2} + 5x + 5x + 1 = 9$ and find the factors $5x(25x + 1) + 1(5x + 1) = 9$
$(5x + 1)(5x + 1) = 9$
Assume that $(5x + 1)(5x + 1)$ are the factors.