Question

Given, $x\ + \ 1\$ is a factor of the polynomial
a).$x^{3} + x^{2}- x + 1$
b).$x^{3} + x^{2} + x + 1$
c).$x^{4} + x^{3} + x^{2} + 1$
d).$x^{4} + 3x^{3} + \ 3x^{2} + x + 1$

Answer 1

In this question, we need to find that $x + 1$ is a factor of which polynomial. By using the factor theorem, we can find the polynomial. Factor theorem is nothing but it links the factors and zeros of the polynomial. If $(k-1)$ is the factor of the polynomial $f(k)$ then $f(1)$ is equal to $0$. This is the factor theorem. In order to find the correct polynomial, we need to check the given polynomials separately.

Complete answer:
Given, the factor $x + 1$
Now we need to find the polynomial for the factor $x + 1$
Let us consider the polynomial as $f(x)$
By factor theorem, if $x + 1$ is the factor of $f(x)$ , then $f( - 1)\ = 0$
a).$\ x^{3} + \ x^{2}- x\ + 1$
Let $f\left( x \right) = \ x^{3} + \ x^{2}\ x\ + 1$
Now we need to find $f( - 1)$ that is we need to substitute $- 1$ in the place of $x$
$f( - 1) = ( - 1)\ + ( - 1)\ - ( - 1)\ + 1$
On simplifying,
We get,
$f( - 1)\ = 2$ which is not equal to 0.
So, $x + 1$ is not the factor of $x^{3} + \ x^{2}- x\ + 1$
b).$x^{3} + \ x^{2} + \ x + 1$
Let $f\left( x \right) = x^{3} + \ x^{2} + \ x + 1$
Now we need to find $f( - 1)$
$f\left( - 1 \right) = \left( - 1 \right)^{3} + \left( - 1 \right)^{2} + \left( - 1 \right) + 1$
On simplifying,
We get,
$f\left( - 1 \right) = 0$
Since $f\left( x \right)$ is equal to $0$ , $x + 1$ is the factor of $x^{3} + \ x^{2} + \ x + 1$
We can also check the other two polynomials also.
c).$x^{4} + \ x^{3} + \ x^{2} + 1$
Let $f\left( x \right) = x^{4} + \ x^{3} + \ x^{2} + 1$
Now we need to find $f( - 1)$
$f\left( - 1 \right) = \left( - 1 \right)^{4} + \left( - 1 \right)^{3} + \left( - 1 \right)^{2} + 1$
On simplifying,
We get,
$f( - 1)\ = 2$ which is not equal to $0$
So $x + 1$ is not the factor of $x^{4} + \ x^{3} + \ x^{2} + 1$
d).$x^{4} + \ 3x^{3} + \ 3x^{2} + \ x\ + 1$
Let $f\left( x \right) = x^{4} + \ 3x^{3} + \ 3x^{2} + \ x\ + 1$
Now need to find $f( - 1)$,
$f( - 1)\ = ( - 1)\ 4 + 3( - 1)\ + 3( - 1)\ + ( - 1)\ + 1$
On simplifying,
We get,
$f\left( - 1 \right) = 1$ which is not equal to $0$
Since $x + 1$ is not the factor of $x^{4} + \ 3x^{3} + \ 3x^{2} + \ x\ + 1$
Hence $\left( x + 1 \right)$ is the factor of $x^{3} + \ x^{2} + \ x + 1$
**Final answer :
$(x + 1)$ is the factor of $x^{3} + \ x^{2}+ \ x + 1$
Option b) $x^{3} + \ x^{2}+ \ x + 1$ is correct. **

Note:
The concept used in this question to find the polynomial factor is factor theorem. The simple example for the factor theorem is if $(x-a)$ is the factor of the polynomial $p(x)$ then $p(a) =0$. It is commonly used to find the roots of the polynomial and also for factoring a polynomial.