Question

How do you factor the trinomial ${{s}^{2}}-4s-5?$

Answer 1

Hint: Trinomial is an expression consisting of three terms. Factorization is the process of writing a polynomial as a product of two or more terms. Usually these terms will leave a remainder of 0 when they divide the actual polynomial as they are the factors of it. We can factorize a given quadratic equation by two ways. Either by splitting the middle term or by directly applying the quadratic equation form and find it’s roots. The quadratic equation formula is :
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ where $\sqrt{{{b}^{2}}-4ac}$ is called the discriminant of the of the quadratic equation.

Complete step by step solution:
Now we have $f\left( x \right)={{s}^{2}}-4s-5$.
Let’s make use of a standard form of a quadratic equation to simplify things.
Let $h\left( x \right)=a{{x}^{2}}+bx+c$
Let’s factorize $f\left( x \right)$ by splitting the middle term.
We split the middle term, $bx$, in such a way so that we write it as the sum of two terms which when multiplied will give us the product of the first, $a{{x}^{2}}$, and last term,$c$, of the quadratic equation.
Upon comparing $f\left( x \right)$ with $h\left( x \right)$ , we conclude :
\[\begin{align}
  & \Rightarrow a=1 \\
 & \Rightarrow b=-4 \\
 & \Rightarrow c=-5 \\
\end{align}\]
Now , let’s split the middle term :
$\begin{align}
  & \Rightarrow f\left( x \right)={{s}^{2}}-4s-5 \\
 & \Rightarrow f\left( x \right)={{s}^{2}}-5s+s-5 \\
\end{align}$
Since $-5s+s=-4s$ and $-5s\times 1s=-5{{s}^{2}}$
$\begin{align}
  & \Rightarrow f\left( x \right)={{s}^{2}}-5s+s-5 \\
 & \Rightarrow f\left( x \right)=s\left( s-5 \right)+1\left( s-5 \right) \\
\end{align}$
Taking $s-5$ common , we get the following :
$\Rightarrow f\left( x \right)=\left( s+1 \right)\left( s-5 \right)$
$\therefore $ Hence we can factor the trinomial$f\left( x \right)={{s}^{2}}-4x-5$ as $\left( s+1 \right)\left( s-5 \right)$ .
We can also try doing it using the formula. As we have already compared and found out the values of $a,b,c$ . We just have to plug it in the formula to get the values of $x$ which are roots to the given quadratic equation. We get the same answer both ways.

Note: We should be very careful while splitting the middle term. And comparing the equation and finding out the coefficients and plugging them in the formula must also be taken care of as they might lead to calculation errors and give wrong factors. If you find splitting the middle term to be difficult, please make use of the formula. If we are trying to factorize any polynomial with degree greater than 2 , we should try to guess at least one of it’s factors by trial and error method. Then divide it with the given polynomial to find all the factors.