How do you factor the trinomial ${n^2} - 10n - 25$?

Hint: First 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.

Formula used: 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:
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 compare the given quadratic equation to the standard quadratic equation and find the value of numbers $a$, $b$ and $c$.
Comparing ${n^2} - 10n - 25$ with $a{x^2} + bx + c$, we get
$a = 1$, $b = - 10$ and $c = - 25$
Now, substitute the values of $a$, $b$ and $c$ in $D = {b^2} - 4ac$ and find the discriminant of the given equation.
$D = {\left( { - 10} \right)^2} - 4\left( 1 \right)\left( { - 25} \right)$
After simplifying the result, we get
$D = 200$
Which means the given equation has real roots.
Now putting the values of $a$, $b$ and $D$ in $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$, we get
$ \Rightarrow x = \dfrac{{10 \pm 10\sqrt 2 }}{{2 \times 1}}$
Divide numerator and denominator by $2$, we get
$ \Rightarrow x = 5 \pm 5\sqrt 2 $
So, $x = 5 + 5\sqrt 2 $ and $x = 5 - 5\sqrt 2 $ are roots of equation ${n^2} - 10n - 25$.

Therefore, the trinomial ${n^2} - 10n - 25$ can be factored as $\left( {n - 5 - 5\sqrt 2 } \right)\left( {n - 5 + 5\sqrt 2 } \right)$.

Note: Given, ${n^2} - 10n - 25$
We have to factor this trinomial.
To factor this trinomial, first we have to find the product of the first and last constant term of the expression.
Here, the first constant term in ${n^2} - 10n - 25$ is $1$, as it is the coefficient of ${n^2}$ and last constant term is $ - 25$, as it is a constant value.
Now, we have to multiply the coefficient of ${n^2}$ with the constant value in ${n^2} - 10n - 25$, i.e., multiply $1$ with $ - 25$.
Multiplying $1$ and $ - 25$, we get
$1 \times \left( { - 25} \right) = - 25$
Now, we have to find the factors of $ - 25$ in such a way that addition or subtraction of those factors is the middle constant term.
Middle constant term or coefficient of $n$ in ${n^2} - 10n - 25$ is $ - 10$.
So, we have to find two factors of $ - 25$, which on multiplying gives $ - 25$ and in addition gives $ - 10$.
We can do this by determining all factors of $25$.
Factors of $25$ are $1,5,25$.
Now among these values find two factors of $ - 25$, which on multiplying gives $ - 25$ and in addition gives $ - 10$.
After observing, we can see that
$\left( { - 5 + 5\sqrt 2 } \right) \times \left( { - 5 - 5\sqrt 2 } \right) = - 25$ and $\left( { - 5 + 5\sqrt 2 } \right) + \left( { - 5 - 5\sqrt 2 } \right) = - 10$
So, these factors are suitable for factorising the given trinomial.
Now, the next step is to split the middle constant term or coefficient of $x$ in these factors.
That is, write $ - 10n$ as $\left( { - 5 + 5\sqrt 2 } \right)n + \left( { - 5 - 5\sqrt 2 } \right)n$ in ${n^2} - 10n - 25$.
After writing $ - 10n$ as $\left( { - 5 + 5\sqrt 2 } \right)n + \left( { - 5 - 5\sqrt 2 } \right)n$ in ${n^2} - 10n - 25$, we get
$ \Rightarrow {n^2} - 10n - 25 = {n^2} + \left( { - 5 + 5\sqrt 2 } \right)n + \left( { - 5 - 5\sqrt 2 } \right)n - 25$
Now, taking $n$ common in ${n^2} + \left( { - 5 + 5\sqrt 2 } \right)n$ and putting in above equation, we get
$ \Rightarrow {n^2} - 10n - 25 = n\left( {n - 5 + 5\sqrt 2 } \right) + \left( { - 5 - 5\sqrt 2 } \right)n - 25$
Now, taking $\left( { - 5 - 5\sqrt 2 } \right)$ common in $\left( { - 5 - 5\sqrt 2 } \right)n - 25$ and putting in above equation, we get
\[ \Rightarrow {n^2} - 10n - 25 = n\left( {n - 5 + 5\sqrt 2 } \right) + \left( { - 5 - 5\sqrt 2 } \right)\left( {n - 5 + 5\sqrt 2 } \right)\]
Now, taking \[\left( {n - 5 + 5\sqrt 2 } \right)\] common in \[n\left( {n - 5 + 5\sqrt 2 } \right) + \left( { - 5 - 5\sqrt 2 } \right)\left( {n - 5 + 5\sqrt 2 } \right)\] and putting in above equation, we get
\[ \Rightarrow {n^2} - 10n - 25 = \left( {n - 5 + 5\sqrt 2 } \right)\left( {n - 5 - 5\sqrt 2 } \right)\]
Therefore, the trinomial ${n^2} - 10n - 25$ can be factored as $\left( {n - 5 - 5\sqrt 2 } \right)\left( {n - 5 + 5\sqrt 2 } \right)$.