Question

How do you simplify $\dfrac{{{x^2} + 4x - 5}}{{{x^2} - 1}}$?

Answer 1

Hint: As the given equation is quadratic in both numerator and denominator in one variable. First, factor the equation in the numerator by factorization. After that apply the formula $\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)$ to factor the denominator. After that cancel out the common factor from the numerator and denominator. So, the simplified term is the desired result.

Complete step-by-step answer:
In mathematics, the number system is the branch that deals with various types of numbers possible to form and easy to operate with different operators such as addition, multiplication, and so on. Another system for variables in mathematics is an algebra system in which we have equations corresponding to some variables through which we can evaluate the values of those variables.
Here, we have one equation to factorize which is involving variable x. Factors are those numbers that completely divide the given number without leaving any remainder. Similarly, factors of an equation will completely divide the equation without leaving any remainder.
We have been given an equation $\dfrac{{{x^2} + 4x - 5}}{{{x^2} - 1}}$.
First, we have to find the factors of the numerator ${x^2} + 4x - 5$ by factorization.
Here, 4 can be written as $\left( {5 - 1} \right)$. Then,
$ \Rightarrow {x^2} + \left( {5 - 1} \right)x - 5$
Open the brackets and multiply the terms,
$ \Rightarrow {x^2} + 5x - x - 5$
Take common factors from the equation,
$ \Rightarrow x\left( {x + 5} \right) - 1\left( {x + 5} \right)$
Take common factors from the equation,
$ \Rightarrow \left( {x + 5} \right)\left( {x - 1} \right)$
So, the factors on the numerator are $\left( {x + 5} \right)\left( {x - 1} \right)$.
Now, we have to find the factors of the denominator ${x^2} - 1$.
We know that,
$\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)$
Use the formula to factorize the denominator,
$ \Rightarrow {x^2} - 1 = \left( {x + 1} \right)\left( {x - 1} \right)$
So, the factors on the denominator are $\left( {x + 1} \right)\left( {x - 1} \right)$.
Now substitute the factored form in numerator and denominator,
$ \Rightarrow \dfrac{{{x^2} + 4x - 5}}{{{x^2} - 1}} = \dfrac{{\left( {x + 5} \right)\left( {x - 1} \right)}}{{\left( {x + 1} \right)\left( {x - 1} \right)}}$
Cancel out the common factors,
$ \Rightarrow \dfrac{{{x^2} + 4x - 5}}{{{x^2} - 1}} = \dfrac{{x + 5}}{{x + 1}}$

Hence, the simplified form of $\dfrac{{{x^2} + 4x - 5}}{{{x^2} - 1}}$ is $\dfrac{{x + 5}}{{x + 1}}$.

Note:
A quadratic equation is a polynomial equation of degree 2. A quadratic equation has two solutions. Either two distinct real solutions, one double real solution, or two imaginary solutions.
There are several methods you can use to solve a quadratic equation:
> Factoring
> Completing the Square
> Quadratic Formula
> Graphing
> All methods start with setting the equation equal to zero.