Question

Determine which of the following polynomials has $(x + 1)$ a factor:
(i)${x^3} + {x^2} + x + 1$
(ii)${x^4} + {x^3} + {x^2} + x + 1$
(iii)${x^4} + 3{x^3} + 3{x^2} + x + 1$
(iv)${x^3} - {x^2} - (2 + \sqrt 2 )x + \sqrt 2 $

Answer 1

Hint: Use factor theorem to determine whether $(x + 1)$ is a factor. Factor theorem states that if $p(a) = 0$ then $(x - a)$ is a factor of $p(x)$.So we need to substitute $x$ by $ - 1$ in each of the polynomials and then check which are turning out to be zero.

Complete step-by-step answer:
The factor $x + 1 = 0 \Rightarrow x = - 1$.
We will now substitute $x = - 1$ in each of the given polynomials and check if $p( - 1) = 0$. If it is equal to zero then $(x + 1)$ is a factor of that polynomial, otherwise it is not.
(i):
Substituting $x = - 1$ in ${x^3} + {x^2} + x + 1$ , we get
$
  {( - 1)^3} + {( - 1)^2} + ( - 1) + 1 \\
   = - 1 + 1 - 1 + 1 \\
   = 2 - 2 \\
   = 0 \\
 $
Since it is equal to $0$ , $(x + 1)$ is a factor of this polynomial.
(ii):
Substituting $x = - 1$ in ${x^4} + {x^3} + {x^2} + x + 1$ , we get,
$
  {( - 1)^4} + {( - 1)^3} + {( - 1)^2} + ( - 1) + 1 \\
   = 1 + ( - 1) + 1 + ( - 1) + 1 \\
   = 1 - 1 + 1 - 1 + 1 \\
   = 3 - 2 \\
   = 1 \\
 $
Since it is not equal to $0$ , $(x + 1)$ is not a factor of this polynomial.
(iii):
Substituting $x = - 1$ in ${x^4} + 3{x^3} + 3{x^2} + x + 1$ , we get,
$
  {( - 1)^4} + 3{( - 1)^3} + 3{( - 1)^2} + ( - 1) + 1 \\
   = - 1 + (3 \times - 1) + (3 \times 1) - 1 + 1 \\
   = - 1 - 3 + 3 - 1 + 1 \\
   = 4 - 5 \\
   = - 1 \\
 $
Since it is not equal to $0$ , $(x + 1)$ is not a factor of this polynomial.
(iv):
Substituting $x = - 1$ in ${x^3} - {x^2} - (2 + \sqrt 2 )x + \sqrt 2 $, we get,
$
  {( - 1)^3} - {( - 1)^2} - (2 + \sqrt 2 )( - 1) + \sqrt 2 \\
   = - 1 - (1) + (2 + \sqrt 2 ) + \sqrt 2 \\
   = - 1 - 1 + 2 + \sqrt 2 + \sqrt 2 \\
   = - 2 + 2 + 2\sqrt 2 \\
   = 2\sqrt 2 \\
 $
Since it is not equal to $0$ , $(x + 1)$ is not a factor of this polynomial.

Note: We could have also verified if $(x + 1)$ is a factor by dividing the polynomials by $(x + 1)$ and checking if the remainder is $0$ but this is an easier method. Another method would have been to factorize the given polynomial and express it in the factor form. From factorization, we could have checked if $x + 1$ is a factor. To do calculations quickly, remember that when $ - 1$ is raised to any odd power, it remains $ - 1$ but when it is raised to any even power, it becomes $1$.