How do you factor $16{x^2} + 8x + 1$?

Hint: First, equate a given polynomial with zero and make it an equation. Next, take $16$ common from the given equation and then divide both sides of the equation by $16$. Next, compare the given quadratic equation to the standard quadratic equation and find the value of numbers $a$, $b$ and $c$ in the given equation. Then, substitute the values of $a$, $b$ and $c$ in the formula of discriminant and find the discriminant of the given equation. Finally, put the values of $a$, $b$ and $D$ in the roots of the quadratic equation formula and get the desired result.
The quantity $D = {b^2} - 4ac$ is known as the discriminant of the equation $a{x^2} + bx + c = 0$ and its roots are given by
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ or $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step by step solution:
First, equate a given polynomial with zero and make it an equation.
$ \Rightarrow 16{x^2} + 8x + 1 = 0$
We know that an equation of the form $a{x^2} + bx + c = 0$, $a,b,c,x \in R$, is called a Real Quadratic Equation.
The numbers $a$, $b$ and $c$ are called the coefficients of the equation.
The quantity $D = {b^2} - 4ac$ is known as the discriminant of the equation $a{x^2} + bx + c = 0$ and its roots are given by
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ or $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
So, first we will take $16$ common from the given equation.
$ \Rightarrow 16\left( {{x^2} + \dfrac{x}{2} + \dfrac{1}{{16}}} \right) = 0$
Divide both sides of the equation by $16$.
$ \Rightarrow {x^2} + \dfrac{x}{2} + \dfrac{1}{{16}} = 0$
Next, compare ${x^2} + \dfrac{x}{2} + \dfrac{1}{{16}} = 0$ quadratic equation to standard quadratic equation and find the value of numbers $a$, $b$ and $c$.
Comparing ${x^2} + \dfrac{x}{2} + \dfrac{1}{{16}} = 0$ with $a{x^2} + bx + c = 0$, we get
$a = 1$, $b = \dfrac{1}{2}$ and $c = \dfrac{1}{{16}}$
Now, substitute the values of $a$, $b$ and $c$ in $D = {b^2} - 4ac$ and find the discriminant of the given equation.
$D = {\left( {\dfrac{1}{2}} \right)^2} - 4\left( 1 \right)\left( {\dfrac{1}{{16}}} \right)$
After simplifying the result, we get
$ \Rightarrow D = 0$
Which means the given equation has real and equal roots.
Now putting the values of $a$, $b$ and $D$ in $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$, we get
$ \Rightarrow x = \dfrac{{ - \dfrac{1}{2} \pm 0}}{{2 \times 1}}$
Divide numerator and denominator by $2$, we get
$ \Rightarrow x = - \dfrac{1}{4}$
$ \Rightarrow 4x = - 1$
$ \Rightarrow 4x + 1 = 0$

Therefore, the trinomial $16{x^2} + 8x + 1$ can be factored as ${\left( {4x + 1} \right)^2}$.

Note: We can also factorize a given trinomial using algebraic identity.
Algebraic identity: ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$
So, rewrite $16{x^2}$ as ${\left( {4x} \right)^2}$.
$ \Rightarrow {\left( {4x} \right)^2} + 8x + 1$
Now, rewrite $1$ as ${1^2}$.
$ \Rightarrow {\left( {4x} \right)^2} + 8x + {1^2}$
Check the middle term by multiplying $2ab$ and compare this result with the middle term in the original expression.
$2ab = 2 \times 4x \times 1$
Now, factor using the perfect square trinomial rule, ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ where $a = 4x$ and $b = 1$.
$ \Rightarrow {\left( {4x + 1} \right)^2}$