Question

Find $\left( x+1 \right)$ is a factor of the polynomial
A. ${{x}^{3}}+{{x}^{2}}x+1$
B. ${{x}^{3}}+{{x}^{2}}+x+1$
C. ${{x}^{4}}+{{x}^{3}}{{x}^{2}}+1$
D. ${{x}^{4}}+{{x}^{3}}+3{{x}^{2}}+x+1$

Answer 1

Hint: Here we have to determine that $\left( x+1 \right)$ is a factor of which of the polynomials given. We will use the factor theorem here which states that $\left( x-a \right)$ is a factor of polynomials $f\left( x \right)$ if $f\left( a \right)=0$ . We will compare the theorem with the factor given and get the value of $a$. Then we will check each option for the value obtained and get our desired answer.

Complete step by step answer:
We have to determine $\left( x+1 \right)$ is a factor of which of the polynomials given using the factor theorem.
Factor theorem states that if we consider a polynomial $f\left( x \right)$ of degree $n\ge 1$ then for any real number $a$ $\left( x-a \right)$ is a factor of $f\left( x \right)$ if $f\left( a \right)=0$ .
The factor is given as,
  $\left( x+1 \right)$
On comparing it with the theorem stated we get,
$\left( x+1 \right)=\left( x-a \right)$
$\Rightarrow \left( x-\left( -1 \right) \right)=\left( x-a \right)$
On comparing we get,
$a=-1$
So we will check which of the options gives the answer as $0$ when the above value is substituted in them.
A. ${{x}^{3}}+{{x}^{2}}x+1$
On substituting $x=-1$ above we get,
$= {{\left( -1 \right)}^{3}}+{{\left( -1 \right)}^{2}}\left( -1 \right)+1$
$= -1+1\times -1+1$
On solving we get,
$= -1-1+1$
$= -1$
As the answer is not $0$ so $\left( x+1 \right)$ is not a factor of ${{x}^{3}}+{{x}^{2}}x+1$ .
B. ${{x}^{3}}+{{x}^{2}}+x+1$
On substituting $x=-1$ above we get,
$= {{\left( -1 \right)}^{3}}+{{\left( -1 \right)}^{2}}+\left( -1 \right)+1$
$= -1+1-1+1$
On solving we get,
$= 0$
As the answer is $0$ so $\left( x+1 \right)$ is the factor of ${{x}^{3}}+{{x}^{2}}+x+1$
C. ${{x}^{4}}+{{x}^{3}}{{x}^{2}}+1$
On substituting $x=-1$ above we get,
$= {{\left( -1 \right)}^{4}}+{{\left( -1 \right)}^{3}}{{\left( -1 \right)}^{2}}+1$
$= 1+\left( -1 \right)\times 1+1$
On solving we get,
$= 1-1+1$
$= 1$
As the answer is not $0$ so $\left( x+1 \right)$ is not a factor of ${{x}^{4}}+{{x}^{3}}{{x}^{2}}+1$ .
D. ${{x}^{4}}+{{x}^{3}}+3{{x}^{2}}+x+1$
On substituting $x=-1$ above we get,
$= {{\left( -1 \right)}^{4}}+{{\left( -1 \right)}^{3}}+3{{\left( -1 \right)}^{2}}+\left( -1 \right)+1$
$= 1-1+3-1+1$
On solving we get,
$= 5-2$
$= 3$
As the answer is not $0$ so $\left( x+1 \right)$ is not a factor of ${{x}^{4}}+{{x}^{3}}+3{{x}^{2}}+x+1$ .
So we get $\left( x+1 \right)$ is a factor of polynomial ${{x}^{3}}+{{x}^{2}}+x+1$

So, the correct answer is “Option B”.

Note:
 Factor Theorem is usually applied in two types of problems: firstly when we have to factorize a polynomial and secondly when we have to find the roots of the polynomial. Another use of it is when we have to remove the known zeros of the polynomial while leaving all unknown zeros intact. It is a special case of a polynomial remainder theorem.