Question

Using the factor theorem, show that $\left( {x - 1} \right)$ is a factor of $\left( {{x^2} - 1} \right)$.

Answer 1

Hint:In the given question, we have to use the factor theorem to prove that the linear polynomial given to us is a factor of the quadratic polynomial function. This question requires us to have the knowledge of basic and simple algebraic rules and operations such as substitution, addition, multiplication, subtraction and many more like these. A thorough understanding of functions, division algorithms and its applications will be of great significance.

Complete step by step answer:
In the given question, we are required to show that $\left( {x - 1} \right)$ is a factor of $\left( {{x^2} - 1} \right)$ using the factor theorem. Factor theorem is a theorem linking the factors of a polynomial with the zeros or roots of the polynomial. It is a special case of the remainder theorem where the remainder on dividing the dividend polynomial by the divisor polynomial comes out to be zero.

According to the factor theorem, if the value of the variable obtained on equating the divisor to zero is a root of the polynomial, then the divisor polynomial is a factor of the dividend polynomial. So, we first equate the divisor polynomial $\left( {x - 1} \right)$ to zero. So, we get the value of variable $x$ as,
$\left( {x - 1} \right) = 0$
$ \Rightarrow x = 1$
Now, substituting the value of $x$ as $1$ in $\left( {{x^2} - 1} \right)$, we get,
${\left( 1 \right)^2} - 1$
$ \Rightarrow 0$
So, the value of the remainder on dividing $\left( {{x^2} - 1} \right)$ by $\left( {x - 1} \right)$ is zero. So, $x = 1$ is a root of the polynomial $\left( {{x^2} - 1} \right)$ and $\left( {x - 1} \right)$ is a factor of $\left( {{x^2} - 1} \right)$ using the factor theorem. Hence, proved.

Note:One must know the significance of the factor theorem in order to solve such problems. Substitution of a variable involves putting a certain value in place of the variable. That specified value may be a certain number or even any other variable. We must take care of the calculations while doing such problems.